SECOND COPY, 
i6b9. 




LIBRARY OF CONGRESS. 



Chap. Copyright No. 

'■ ShelfXGk_Zl 
~T+V^L 

UNITED STATES OF AMERICA. 




ELEMENTS OF GRAPHIC STATICS. 



THE 



ELEMENTS OF GRAPHIC STATICS 



^ ^cxt'Moo'k for Stutrtnts of lEnsinccrmg 



L. M. HOSKINS 

Professor of Applied Mechanics in the Leland Stanford 
Junior University 



REVISED EDITION 



THE MACMILLAN COMPANY 

LONDON : MACMILLAN & CO., Ltd. 
1899 

Aii rights reserved 






24054 



Copyright, 1892, 
By MACMILLAN AND CO. 



Copyright, 1899, 
By the MACMILLAN COMPANY. 



Set up and electrotyped August, 1892. New edition, revised, 
printed from new plates January, 1899. 

TWO COPIES REC-iVED. 




J. S. Gushing & Co. — Berwick & Smith 
Norwood Mass. U.S.A. 






"jCj^H 



PREFACE TO REVISED EDITION. 

The method of treatment adopted in this work is designed 
to meet the needs of the beginner. To this end the endeavor 
has been made to secure simplicity of presentation without 
sacrifice of logical rigor. 

In scope, the work has been planned with reference to the 
requirements of students of engineering. This limits the de- 
velopment of the general theory to such principles and methods 
as are practically useful. It also excludes many applications 
which, though leading to practical results, are likely to prove 
useful and to save labor only in the hands of the expert in 
graphical methods. Graphic Statics is treated as a branch of 
Mechanics rather than of Geometry, and those beautiful devel- 
opments whose chief interest is geometrical have not been 
included. 

Although graphical methods are especially useful to the 
structural engineer, it is believed that students in all depart- 
ments of engineering will find it profitable to become familiar 
with the general theory of complanar forces from the graphical 
side, as well as with the simpler applications to the determina- 
tion of stresses in framed structures, and of centroids and mo- 
ments of inertia of plane areas. The application to trusses 
carrying moving loads is of less general interest, its practical 
utility being limited mainly to the solution of problems in 
bridge design. The methods developed in Chapter VII will 
therefore be of practical value mainly to students giving special 
attention to this branch of engineering. 

In the present revised edition no change has been made in 
general plan, and few changes in the treatment adopted, except 



vi PREFACE TO THE REVISED EDITION. 

in the portions relating to beams and trusses carrying moving 
loads (Chapters VI and VII). These portions have been 
wholly re-written. It is believed that a substantial improve- 
ment has been made upon the methods hitherto used, par- 
ticularly in the criterion for determining the position of a 
given load-series which causes maximum stress in any member 
of a truss. The improvement consists in generalization, which 
is believed to be gained without sacrifice of simplicity. The 
graphical method of applying the criterion in the case of 
trusses with parallel chords has been fully treated by Pro- 
fessor H. T. Eddy. The method here given applies without 
the restriction to parallel chords. The algebraic statement of 
the same criterion, as given in Art. 152, is also believed to 
be a useful generalization of the methods heretofore used. 
Whether the algebraic or the graphical treatment is preferred, 
a method is useful in proportion to its generality, provided this 
does not involve a loss of simplicity. There is a decided ad- 
vantage in the use of a single general equation, applicable to 
any member of any truss, instead of several particular equa- 
tions, each applicable to a special member or to a special form 
of truss. 

Stanford University, California, 
November, 1898. 



CONTENTS. 



PART I. — GENERAL THEORY. 
Chapter I. Definitions. — Concurrent Forces. 

PAGE 

§ I. Preliminary Definitions i 

2. Composition of Concurrent Forces 5 

3. Equilibrimn of Concurrent Forces 6 

4. Resolution of Concurrent Forces 8 

Chapter II Non-concurrent Forces. 

§ I. Composition of Ncn-concurrent Forces Acting on the Same Rigid 

Body II 

2. Equilibrium of Non-concurrent Forces 18 

3. Resolution into Non-concurrent Systems 28 

4. Moments of Forces and of Couples 30 

5. Graphic Determination of Moments 35 

6. Summary of Conditions of Equilibrium 37 

Chapter III. Internal Forces and Stresses. 

§ I. External and Internal Forces 40 

2. External and Internal Stresses 42 

3. Determination of Internal Stresses 46 

PART II. — STRESSES IN SIMPLE STRUCTURES. 

Chapter IV. Introductory. 

§ I . Outline of Principles and Methods 53 



viii CONTENTS. 



Chapter V. Roof Trusses. — Framed Structures Sustaining 
Stationary Loads. 

PAGE 

§ I. Loads on Roof Trusses 58 



Roof Truss with Vertical Loads 61 

Stresses Due to Wind Pressure 67 

Maximum Stresses 71 

Cases Apparently Indeterminate 74 

Three-hinged Arch . 80 

Counterbracins: 88 



Chapter VI. Simple Beams. 

§ I. General Principles 94 

2. Beam Sustaining Fixed Loads 98 

3. Beam Sustaining Moving Loads loi 

Chapter VII. Trusses Sustaining Moving Loads. 

§ I. Bridge Loads 112 

2. Truss Regarded as a Beam .115 

3. Truss Sustaining Any Series of Moving Loads 120 

4. Truss with Subordinate Bracing 137 

5. Uniformly Distributed Moving Load 145 

PART III. — CENTROIDS AND MOMENTS OF INERTIA. 

Chapter VIII. Centroids. 

§ I. Centroid of Parallel Forces 152 

2. Center of Gravity — Definitions and General Principles .... 157 

3. Centroids of Lines and of Areas 160 

Chapter IX. Moments of Inertia. 

§ I. Moments of Inertia of Forces 167 

2. Moments of Inertia of Plane Areas 177 

Chapter X. Curves of Inertia. 

§ I. General Principles 187 

2. Inertia-Ellipses for Systems of Forces 190 

3. Inertia-Curves for Plane Areas 195 



GRAPHIC STATICS. 

Part I. 
GENERAL THEORY. 



CHAPTER I. — DEFINITIONS. CONCURRENT 
FORCES. 

§ I. Preliminary Definitions. 

1. Dynamics treats of the action of forces upon bodies. Its 
two main branches are Statics and Kinetics. 

Statics treats of the action of forces under such conditions 
that no change of motion is produced in the bodies acted upon. 

Kinetics treats of the laws governing the production of motion 
by forces. 

2. Graphic Statics has for its object the deduction of the 
principles of statics, and the solution of its problems, by means 
of geometrical figures. 

3. A Force is that which tends to change the state of motion 
of a body. We conceive of a force as a push or a pull applied 
to a body at a definite point and in a definite direction. Such 
a push or pull tends to give motion to the body, but this ten- 
dency may be neutralized by the action of other forces. The 
effect of a force is completely determined when three things 
are given, — its magnitude, its direction, and its point of applica- 
tion. The line parallel to the direction of the force and con- 
taining its point of application, is called its lijie of action. 



2 GRAPHIC STATICS. 

Every force acting upon a body is exerted by some other body. 
But the problems of statics usually concern only the body acted 
upon. Hence, frequently, no reference is made to the bodies 
exerting the forces. 

4. Unit Force. — The iinit force is a force of arbitrarily chosen 
magnitude, in terms of which forces are expressed. Several 
different units are in use. The one employed in this work is 
X\\Q. pound, which will now be defined. 

A pound force is a force equal to the weight of a pound mass 
at the earth's surface. A pound mass is the quantity of matter 
contained in a certain piece of platinum, arbitrarily chosen, and 
established as the standard by act of the British Parliament. 

The pound force, as thus defined, is not perfectly definite, 
since the weight of any given mass (that is, the attraction of 
the earth upon it) is not the same for all positions on the earth's 
surface. The variations are, however, unimportant for most of 
the requirements of the engineer. 

In its fundamental meaning, the word "pound" refers to the 
unit mass, and it is unfortunate that it is also applied to the unit 
force. The usage is, however, so firmly established that it will 
be here followed. 

5. Concurrent and Non-concurrent Forces. — Forces acting on 
the same body are concurrent when they have the same point of 
application. When applied at different points they are non- 
concurrent. 

6. Complanar Forces are those whose lines of action are in 
the same plane. In this work, only complanar systems will be 
considered unless otherwise specified. 

7. A Couple is the name given to a system consisting of two 
forces, equal in magnitude, but opposite in direction, and having 
different lines of action. The perpendicular distance between 
the two lines of action is called the arm of the couple. 



PRELIMINARY DEFINITIONS. 3 

8. Equivalent Systems of Forces. — Two systems of forces 
are equivalent when either may be substituted for the other 
without change of effect. 

9. Resultant. — A single force that is equivalent to a given 
system of forces is called the resultant of that system. It will 
be shown subsequently that a system of forces may not be 
equivalent to any single force. When such is the case, the 
simplest system equivalent to the given system may be called 
its resultant. Any forces having a given force for their 
resultant are called components of that force. 

10. Composition and Resolution of Forces. — Having given 
any system of forces, the process of finding an equivalent system 
is called the composition of forces, if the system determined con- 
tains fewer forces than the given system. If the reverse is the 
case, the process is called the resolution of forces. 

The process of finding the resultant of any given forces is 
the most important case of composition ; while the process of 
finding two or more forces, which together are equivalent to a 
single given force, is the most common case of resolution. 

11. Representation of Forces Graphically. — The magnitude 
and direction of a force can both be represented by a line ; the 
length of the line representing the magnitude of the force, and 
its direction the direction of the force. 

In order that the length of a line may represent the magni- 
tude of a force, a certain length must be chosen to denote the 
unit force. Then a force of any magnitude will be represented 
by a length which contains the assumed length as many times 
as the magnitude of the given force contains that of the unit 
force. 

In order that the direction of a force may be represented by 
a line, there must be some means of distinguishing between the 
two opposite directions along the line. The usual method is to 
place an arrow-head on the line, pointing in the direction toward 



4 GRAPHIC STATICS. 

which the force acts. If the line is designated by letters placed 
at its extremities, the order in which these are read may indicate 
the direction of the force. Thus, AB and BA represent two 
forces, equal in magnitude but opposite in direction. 

The /me of action of a force can also be represented by a line 
drawn on the paper. 

In solving problems in statics, it is usually convenient to 
draw two separate figures, in one of which the forces are 
represented in magnitude and direction only, and in the other 
in line of action only. 

These two species of diagrams will be called force diagrams 
and space diagrams, respectively. 

12. Notation. — The use of graphic methods is much facili- 
tated by the adoption of a convenient system of notation in 
the figures drawn. 

There will generally be two figures (the force diagram and 
the space diagram) so related that for every line in one there 
is a corresponding line in the other. 

In the force diagram each line represents a force in magni- 
tude and direction ; in the space diagram the corresponding 
line represents the line of action of the force. These lines 
will usually be designated in the ,following manner : The line 
denoting the magnitude and direction of 
the force will be marked by two capital 
letters, one at each extremity ; while the 
action-line will be marked by the corre- 
sponding small letters, one being placed at 
each side of the line designated. Thus, in 
Fig. I, AB represents a force in magni- 
tude and direction, while its action-line is marked by the letters 
ab, placed as shown. 




COMPOSITION OF CONCURRENT FORCES. 




Ei 



The 



§ 2. Composition of Conciirrejit Forces. 

13. Resultant of Two Concurrent Forces. — If two concur- 
rent forces are represented in magnitude and direction by two 
lines AB and BC, their 
resultant is represented 
in magnitude and direc- 
tion by AC. (Fig. 2.) 
Proofs of this proposi- 
tion are given in all 
elementary treatises on 

mechanics, and the demonstration will be here omitted. 
point of application of the resultant is the same as that of the 
given forces. Thus if O (Fig. 2) is the given point of applica- 
tion, then ab, be, and ac, drawn parallel to AB, BC, and AC 
respectively, are the lines of action of the two given forces and 
their resultant. The figure marked i^A) is deforce diagram, and 
{B) is the corresponding space diagram (Art. 11). 

14. Resultant of Any Number of Concurrent Forces. — If any 

number of concurrent forces are represented in magnitude and 
direction by lines AB, BC, CD, . . ., their resultant is repre- 
sented in magnitude and direction by the line AN, where N is 
the extremity of the line representing the last of the given 
forces. 

This proposition follows immediately from the preceding 
one ; for the resultant of the forces represented by AB and 
BC is a force repre- 
sented hy AC; the re- 
sultant oi AC and C-D 
is AD, and so on. By 
continuing the process 
we shall arrive at the 
result stated. It is 

readily seen that the order in which the forces are taken does 
not affect the magnitude or direction of the resultant as thus 






6 GRAPHIC STATICS. 

determined. The point of application of the resultant is the 
same as that of the given forces. Figure 3 shows the force 
diagram and space diagram for a system of four forces repre- 
sented by AB, BC, CD, DE, and their resultant represented 
by AE, applied at the point O. 

From the above construction it is evident that every system 
of concurrent forces has for its resultant some single force 
(Art. 9) ; though in particular cases its magnitude may be zero. 

15. Force Polygon. — The figure formed by drawing in suc- 
cession lines representing in magnitude and direction any 

number of forces is called a force polygon for 

those forces. Thus Fig. 4 is a force polygon 

for any four forces represented in magnitude 

and direction by the lines AB, BC, CD, and 

DE, whatever their lines of action. It may 

■^^°' happen that the point E coincides with A, in 

which case the polygon is said to be closed. It is evident that 

the order in which the forces are taken does not affect the 

relative positions of the initial and final points. 

§ 3. Eqidlibriiim of Concurrent Forces. 

16. Definition. — A system of forces acting on a body is 
in eqnilibrmni if the motion of the body is unchanged by its 
action. 

1 7. Condition of Equilibrium. — In order that no motion 
may result from the action of any system of concurrent forces, 
the magnitude of the resultant must be zero ; and conversely, 
if the magnitude of the resultant is zero, no motion can 
result. But (Arts. 14 and 15) the condition that the result- 
ant is zero is identical with the condition that the force 
polygon closes. Hence, the following proposition : 

If any system of concurrent forces is in equilibrinin, the force 
polygon for the system must close. And conversely, If the force 



EQUILIBRIUM OF CONCURRENT FORCES. 



7 





Fig 



polygon is closed for any system of conacn-ent forces, the system 
is in eqidlibriimi. 

The comparison of this with the analytical conditions of 
equilibrium is given in Art. 22. 

18. Method of Solving Problems in Equilibrium. — If a sys- 
tem of concurrent forces in equilibrium be partially unknown, 
we may in certain cases determine the unknown elements by 
applying the principles of Art. 17. 

A common case is that in which two forces are unknown 
except as to lines of action. Thus, suppose a system of five 
forces in equilibrium, 
three being fully 
known, represented in 
magnitude and direc- 
tion by AB, BC, CD 
(Fig. 5), and in lines 
of action by ab, be, cd, 

while concerning the other two we know only their lines of 
action ' de, ea. To determine these two in magnitude and 
direction, it is necessary only to complete the force polygon of 
which ABCD is the known part. The remaining sides must be 
parallel respectively to de and ea. From D draw a line parallel 
to de, and from A a line parallel to ea, prolonging them till they 
intersect at E. Then DE and EA represent the required forces 
in magnitude and direction, and the complete force polygon is 
ABCDEA. It is evident that ABCDE'A is an equally legiti- 
mate form of the force polygon, and gives the same result for 
the magnitude and direction of each of the unknown forces. 
This problem occurs constantly in the construction of stress 
diagrams by the method described in Part II. 

The student will find little difficulty in treating other prob- 
lems in the equilibrium of concurrent forces. 

19. Problems in Equilibrium. — (i) A particle is in equilib- 
rium under the action of five forces, three of which are 



8 GRAPHIC STATICS. 

completely known, while of the remaining two, one is known 
in direction only, and the other in magnitude only. To deter- 
mine the unknown forces. 

(2) Suppose two forces known in magnitude but not in 
direction, the remaining forces being wholly known. 

(3) Suppose one force wholly unknown, the others being 
known. 

§ 4. Resolution of ConciLrrent Forces. 

20. To Resolve a Given Force into Any Number of Com- 
ponents having the same point of application, we have only to 
draw a closed polygon of which one side shall represent the 
magnitude and direction of the given force ; then the remaining 
sides will represent, in magnitude and direction, the required 
components. This problem is, in general, indeterminate, unless 
the components are required to satisfy certain specified con- 
ditions. 

[Note. — A problem is said to be indeterminate if its conditions can be satisfied 
in an infinite number of ways. It is deterniinate if it admits of only one solution. 
Thus, the problem, to determine the values of x and y which shall satisfy the equation 
X -\- y ^ 10, is indeterminate; while the problem, given 2jc+ 3= 7) to find the 
value of j;, is determinate. The case in which a finite number of solutions is possible 
may be called incompletely dete^'tninate. Thus, the problem, given ^^ -)- ^ — 6 = O, 
to find X, admits of two solutions, and therefore is incompletely determinate. All 
these classes of problems may be met with in statics.] 

21. To Resolve a Given Force into Two Components. — This 
problem is indeterminate unless additional data are given. For 
if the given force be represented in magnitude and direction by 
a line, any two lines which with the given line form a triangle 
may represent forces which are together equivalent to the given 
force. But an infinite number of such triangles may be drawn. 
The solution of the following four cases of this problem will 
form exercises for the student. In each case the force diagram 
and space diagram should be completely drawn, and the student 
should notice whether the problem is determinate, partially 
determinate, or indeterminate. 



RESOLUTION OF CONCURRENT FORCES. 



(i) Let the lines of action of the required components be 
given. 

(2) Let the two components be given in magnitude only. 

(3) Let the line of action of one component and the magni- 
tude of the other be given. 

(4) Let the magnitude and direction of one component be 
given. 

It will be noticed that these four cases correspond to four 
cases of the solution of a plane triangle. 

22. Resolved Part of a Force. — If a force is conceived to be 
replaced by two components at right angles to each other, each 
is called the resolved part,^' in its direction, of the given force. 

It is readily seen that the resolved part of a force repre- 
sented hy AB (Fig. 6) in the direction of any line XX, is 
represented in magnitude and direction » 

by A'B', the orthographic projection of 
AB upon XX. It follows that the 
resolved part (in any given direction) of 
the resultant of any concurrent forces 
is equal to the algebraic sum of the 
resolved parts of its components in that direction ; signs plus 
and minus being given to the resolved parts to distinguish the 
two opposite directions which they 
may have. Thus (Fig. 7) the re- 
solved parts of the forces AB, BC, 
CD, in a direction parallel to XX, 
are A'B', B'C, CD' ; and their 
algebraic sum is A'D', which is the 
resolved part of the resultant AD. 
If the resultant is zero, D' coincides with A' ; hence, 

(i) For the equilibrium of any concurrent forces, the sum 
•of their resolved parts in any direction must be zero. 





Fig. v 



* The term "resolute" has been proposed by J. B. Lock (" Elementary Statics ") to 
denote what is here defined as the resolved part of a force. 



lO GRAPHIC STATICS. 

Again, if D^ coincides with A\ then either D coincides 
with A, or else AD is perpendicular to XX \ hence, 

(2) If the sum of the resolved parts of any concurrent 
forces in a given direction is zero, their resultant (if any) 
is perpendicular to that direction. And if the sum of the 
resolved parts is zero for each of two directions, the resultant 
is zero, and the system is in equilibrium. 

Propositions (i) and (2) state the conditions of equilibrium 
for concurrent forces usually deduced in treatises employing 
algebraic methods. 



CHAPTER II. — NON-CONCURRENT FORCES. 

§ I. Composition of No7i-co7icitrrent Forces Acting on the Same 

Rigid Body. 

23. Definition of Rigid Body. — A rigid body is one whose 
particles do not change their positions relative to each other 
under any applied forces. No known body is perfectly rigid, 
but for the purposes of statics, most solid bodies may be con- 
sidered as such ; and any body which has assumed a form of 
equilibrium under applied forces, may, for the purposes of 
statics, be treated as a rigid body without error. 

24. Change of Point of Application. — The effect of a force 
upon a rigid body will be the same, at whatever point in its 
line of action it is applied, if only the particle upon which it 
acts is rigidly connected with the body. 

This proposition is fundamental to the development of the 
principles of statics, and is amply justified by experience.* 
In applying the principle, we are at liberty to assume a 
point of application outside the actual body, the latter being 
ideally extended to any desired limits. 

25. Resultant of Two Non-Parallel Forces. — If two com- 
planar forces are not parallel, their lines of action must inter- 
sect, and the point of intersection may be taken as the point 
of application of each force. Hence, they may be treated as 



* This proposition may be proved analytically by deducing the equations of motion 
of a rigid body, and showing that the effect of any force on the motion of the body 
depends only upon its magnitude, direction, and line of action. But such a proof is, of 
course, outside the scope of this work. 

II 



12 GRAPHIC STATICS. 

concurrent forces, and their resultant may be determined as 
in Art. 13. The following proposition may therefore be 
stated : 

If two forces acting in the same plane on a rigid body 
are represented in magnitude and direction by AB arid BC, 
their resultant is represented in magnitude and direction by 

AC, and its line of action passes through the point of inter- 
section of the lines of action of the given forces. Its point 
of application may be any point of this line. 

It may happen that the point of intersection of the two 
given lines of action falls outside the limits available for the 
drawing. In such a case it will be most convenient to find 
the resultant by the method to be explained in Art. 27. The 
same remark applies to the case of two parallel forces. 

26. Resultant of Any Number of Non-concurrent Forces — 
First Method. — The method of the preceding article may be 
extended to the determination of the resultant of any number 
of forces acting on the same rigid body. Let AB, BC, CD, DE 

(Fig. 8), represent in 
magnitude and direction 
four forces, and let ab, 
be, cd, de represent their 
lines of action. To find 
their resultant, we may 
proceed as follows : 
The resultant of AB 
and BC is represented 
in magnitude and direction by AC, and in line of action by ac 
drawn parallel to AC through the point of intersection of ab 
and be. Combining this resultant with CD, we get as their 
resultant a force represented in magnitude and direction by 

AD, and in line of action by ad drawn parallel' to AD through 
the point of intersection of ac and cd. This is evidently the 
resultant of AB, BC, and CD. In the same way, this resultant 




COMPOSITION OF NON-CONCURRENT FORCES. 



13 



combined with DE gives for their resultant a force whose mag- 
nitude and direction are represented by AE^ and whose hne of 
action is ae, parallel to AE and passing through the point in 
which ad intersects de. This last force is the resultant of the 
four given forces. 

The process may evidently be extended to the case of any 
number of forces. 

As in the case discussed in the preceding article, this method 
will become inapplicable or inconvenient if any of the points of 
intersection fall outside the limits available for the drawing. 
For this reason it is usually most convenient to employ the 
method described in Art. 27. 

The student should bear in mind that the length and direc- 
tion AE and the line ae are not the magnitude, direction, and 
line of action of any actual force applied to the body. By the 
resultant is meant an ideal force, which, if it acted, would 
produce the same effect upon the motion of the body as is 
produced by the given forces. It is a force which may be 
conceived to replace the actual forces, and may be assumed to 
be applied to any particle in its line of action, provided that 
particle is regarded as rigidly connected with the given body. 
The line of action may in reality fail to meet the given body. 
(See Art. 24.) 

27. Resultant of Non-concurrent Forces — Second Method. — 

This method will be described by reference to an example. 
Referring to Fig. 9, let AB, BC, CD, DE represent in magni- 
tude and direction four forces whose lines of action are ab, be, 
cd, de ; and let it be required to find their resultant. Draw the 
force polygon ABCDE, and from any point O in the force 
diagram draw lines OA, OB, OC, OD, OE. These lines may 
represent, in magnitude and direction, components into which 
the given forces may be resolved. Thus AB is equivalent to 
forces represented hy AO and OB acting in any lines parallel 
to AO, OB, whose point of intersection falls upon ab ; BC is 



14 



GRAPHIC STATICS. 



equivalent to forces represented by BO, OC, acting in any lines 
parallel to BO, OC, which intersect on be ; and so for each of 
the given forces. The four given forces may, therefore, be 
replaced by eight forces given in magnitude and direction by 




AO, OB, BO, OC, CO, OD, DO, OE, with proper lines of 
action. Now, it is possible to make the lines of action of the 
forces represented by OB and BO coincide ; and the same is 
true of each pair of equal and opposite forces, OC, CO ; OD, 
DO. To accomplish this, let AO, OB act in lines ao, ob, inter- 
secting at any assumed point of ab. Prolong ob to intersect 
be, and take the point thus determined as the point of inter- 
section of the lines of action of BO, OC ; these lines are then 
bo, oe. Similarly prolong oc to intersect ed, and let the point 
of intersection be taken as the point at which CD is resolved 
into CO and OD ; the lines of action of these forces are then 
CO and od. In like manner choose do, oe, intersecting on de, as 
the lines of action of DO, OE. If this is done, the forces OB, 
BO will neutralize each other and may be omitted from the 
system ; also the pairs OC, CO, and OD, DO. Hence, there 
remain only the two forces represented in magnitude and 
direction by AO, OE, and in lines of action by ao, oe. 
Their resultant is given in magnitude and direction by AE, 
and its line of action is ae, drawn parallel to AE through the 



COMPOSITION OF NON-CONCURRENT FORCES. 



15 



point of intersection of oa and oe ; and this is also the result- 
ant of the given system. 

By carefully following through this construction the student 
will be able to reduce it to a mechanical method, which can 
te readily applied to any system. 

2Z. Funicular Polygon. — The polygon whose sides are oa, ob, 
■oc, od, oe, is called z. funicular polygoji * for the given forces. 

Since the point at which the two components of AB are 
assumed to act may be taken anywhere on the line ab, there 
may be any number of funicular polygons with sides parallel 
to oa, ob, etc. Again, if O is taken at a different point, there 
may be drawn a new funicular polygon starting at any point 
of ab ; and by changing the starting point any number of 
funicular polygons may be drawn with sides parallel to the 
new directions of OA, OB, etc. Moreover, different force 
and funicular polygons may be obtained by changing the order 
in which the forces are taken. 

It may be proved geometrically that for every possible funic- 
ular polygon drawn for the same system of forces, the last 
vertex, determined by the above method, will lie on the same 
line parallel to the closing side of the force polygon (as ae. 
Fig. 9). Such a proof is outside the scope of this work. 
The truth of the proposition may, however, be shown from 
the principles of statics. For if it were not true, it would 
be possible by the above method to find two or more forces, 
having dijf event lines of action, which are equivalent to each 
other, because each is equivalent to the given system. But 
this is impossible. 

29. Examples. — i. Choose five forces, assigning the magni- 
tude, direction, and line of action of each, and find their 
resultant by constructing the force and funicular polygons. 

2. Draw a second funicular polygon, using the same point 
O in the force diagram. 

*Also called equilibrium polygon. 



l6 GRAPHIC STATICS. 

3. Draw a third funicular polygon, choosing a new point O^ 

4. Solve the same problem, taking the forces in a different 
order. 

30. Definitions. — The point O (Fig. 9) is called the pole of 
the force polygon. The lines drawn from the pole to the 
vertices of the force polygon may be called rays. The sides 
of the funicular polygon are sometimes called strings. 

Each ray in the force diagram is parallel to a corresponding" 
string in the space diagram. As a mechanical rule, it should 
be remembered that the two rays drawn to the extremities of the 
line representing any force are respectively parallel to the twO' 
strings which intersect on the line of action of that force. 

The rays terminating at the extremities of any side of the 
force polygon represent in magnitude and direction two com- 
ponents that may replace the force represented by that side ; 
while the corresponding strings represent the lines of action of 
these components. Thus (Fig. 9) the force represented by BC 
may be replaced by two forces represented by BO, OC, acting 
in the lines bo, oc. 

The pole distance of any force is the perpendicular distance 
from the pole to the line representing that force in the force 
diagram. It may evidently be considered as representing the 
resolved part, perpendicular to the given force, of either of the 
components represented by the corresponding rays. Thus, OM 
(Fig. 9) is the pole distance of AB ; and OM represents the 
resolved part, perpendicular to AB, of either OA or OB. The 
student should notice particularly that the pole distance repre- 
sents 2. force magnitude and not a length. 

31. Forces not Possessing a Single Resultant. — It may hap- 
pen that the first and last sides of the funicular polygon are 
parallel, so that the above construction fails to give the line of 
action of the resultant. This will be the case if the pole is 
chosen on the line AE (Fig. 9), because the first and last sides 
of the funicular polygon are respectively parallel to OA and 



COMPOSITION OF NON-CONCURRENT FORCES. 17 

OE. The difficulty will, in this case, be avoided by taking the 
pole at some point not on the line AE. But in one particular 
case OA and OE will be parallel, wherever the pole be taken. 
By inspection of the force diagram it is seen that this can 
occur only when the point E coincides with A. In this case 
AO and OE represent equal and opposite forces; and unless 
their lines of action coincide, they cannot be combined into a 
simpler system. If their lines of action are coincident, the 
forces neutralize each other and the resultant is zero ; if not, 
the system reduces to a couple (Art. 7). If the lines of action 
oi AO and OE coincide, they may still be regarded as forming 
a couple, its arm being zero. Therefore, 

If tJie force polygon for any system of forces closes, the result- 
ant is a couple. 

Now, it is evident that by shifting the starting point for the 
funicular polygon, the lines of action oi AO and OE will be 
shifted ; and by taking a new pole, their magnitude, or direc- 
tion, or both, may be changed. Hence, there may be found 
any number of couples, each equivalent to the same system 
of forces, and therefore equivalent to each other. The relation 
between equivalent couples is discussed in Art. 52. 

Example. — Assume five forces whose force polygon closes, 
the lines of action being taken at random. Draw two funicular 
polygons, using the same pole, and a third, using a different 
pole ; thus reducing the given system to an equivalent couple 
in three different ways. 

32. Resultant Force and Resultant Couple. — From the above 
discussion (Arts. 27 and 31), it is evident that any system of 
complanar forces, acting on the same rigid body, is equivalent 
either to a single force or to a couple. In other words, every 
system of complanar forces possesses either a resultant force or a 
resultant couple. (See Art. 9.) 

33. Comparison of Methods. — Of the methods given in Arts. 
26 and 2^, for finding the resultant of a system of non-concur-^ 



1 8 GRAPHIC STATICS. 

rent forces, the first is a special case of the second. For if the 
pole in Fig. 9 be chosen at the point A, and the first string be 
made to pass through the point of intersection of ab and be, the 
construction becomes identical with that of Fig. 8. If the first 
method is employed, we are liable to meet the difficulty men- 
tioned in Arts. 25 and 26, that some of the required points of 
intersection do not fall within convenient limits. 

In the second method, since the pole may be chosen at pleas- 
ure, the rays may generally be caused to make convenient angles 
with the given forces, and all required points of intersection to 
fall within convenient limits. 

34. Closing of the Funicular Polygon. — Let the force poly- 
gon be drawn for any given forces, and let A and E be the 
initial and final points. Suppose a funicular polygon drawn, 
corresponding to any pole O. The given system is equivalent 
to two forces represented in magnitude and direction by AO, 
OE, their lines of action being the first and last sides of the 
funicular polygon. In general, these lines of action will not be 
parallel; but, as explained in Art. 31, it may happen that they 
■are parallel. And if parallel, it may happen that they coincide. 
In this case, the funicular polygon is said to be elosed. 

§ 2. Eqiiilibrmin of Non-co7ieurrent Forees. 

35. Conditions of Equilibrium. — From the foregoing princi- 
ples the conditions of equilibrium for any system of complanar 
forces may be deduced. 

Whatever the system of forces may be, let a force polygon be 
drawn, and let A and E be its initial and final points. Choose 
a pole O, and draw a funicular polygon. The system is thus 
reduced to two forces represented in magnitude and direction 
loy AO and OE ; their lines of action being the first and last 
sides of the funicular polygon. In order that there may be 
equilibrium, these two forces must be equal and opposite and 
have the same line of action. In order that they may be equal 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 19 

and opposite, A and E must coincide. In order that they may- 
have the same line of action, the first and last sides of the funic- 
ular polygon must coincide. That is, in order that there may 
be equilibrium, two conditions must be satisfied : 

(a) TJie force polyg07i must close. 

{b) The fiinicjdar polygon must close. 

Conversely, if the force polygon closes mid one funicular poly- 
gon also closes, the system is in egnilibriiim. For, if the force 
polygon closes, the two forces AO and OE are equal and oppo- 
site; and if a funicular polygon closes, these two forces have 
the same line of action and therefore balance each other. 

It follows that if the force polygon and one funicular polygon 
are closed, all funicular polygons will close. 

36. Auxiliary Conditions of Equilibrium. — If any system of 
forces in equilibrium be divided into two groups, the resultants 
of the two groups are equal and opposite and have the same 
line of action. 

Particular case. — If three forces are in equilibrium, their 
lines of action must meet in a point, or be parallel. For, if 
the lines of action of two of the forces intersect, their result- 
ant must act in a line passing through the point of inter- 
section. But this resultant must be equal and opposite to 
the third force and have the same line of action. 

37. General Method of Solving Problems in Equilibrium. — 

The principles of Art. 35 furnish a general method of solv- 
ing, graphically, problems relating to the equilibrium of com- 
planar forces acting on a rigid body. The problems to be 
solved will always be of the following kind : 

A body is in equilibrium under the action of forces, some 
of which are completely known, and others wholly or partly 
unknown. It is required to determine the unknown elements. 
The required elements may be either the magnitudes or the 
lines of action of the forces. 



20 



GRAPHIC STATICS. 



The general method of procedure is as follows : First, drawr 
the force polygon and funicular polygon so far as possible 
from the given data ; then complete them, subject to the 
condition that both polygons must close. 

This general method will be illustrated by the solution of 
several problems of frequent occurrence. We may here meet 
with both determinate and indeterminate problems (Art. 20). 
In order that a problem may be determinate, the given data, 
together with the condition that the force polygon and funic- 
ular polygon are to close, must be just sufficient to fully 
determine these figures. There are many possible cases fur- 
nishing determinate problems. Some of these cannot readily 
be solved graphically. In the following articles are treated 
three important cases to which the general method above 
outlined is well adapted. 



a d 



38, Problems in Equilibrium. I. — A rigid body is in equi- 
librium under the action of a system of parallel forces, all 
known except two, these being unknown in magnitude and 
direction, but having known lines of action. It is required 
to fully determine the unknown forces. 

Let the known lines of action of five forces in equilibrium 

be ab, be, cd, de, ea (Fig. 10), and let 
AB, BC, CD represent the known 
forces in magnitude and direction, 
while DE and EA are at first un- 
known. The force polygon, so far 
as it can be drawn from the given 
data, is the straight line ABCD. 
Choose a pole O and draw rays OAy 
OB, OC, OD. Parallel to these draw 
in the space diagram the strings oa^ 
ob, OC, od, four successive sides of the 
funicular polygon. This much can 
be drawn from the data given. We must now complete the 



Fig. 10 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 2 1 

two polygons and cause both to close. In the funicular poly- 
gon, but one side remains to be drawn ; and in the force 
polygon but one vertex remains unknown. If the forces are 
taken in the order AB, BC, CD, DE, EA, the unknown vertex 
in the force polygon is the one to be marked E ; the unknown 
side of the funicular polygon is oe, and is to be parallel to 
the line OE. But the string oe must pass through the inter- 
section of od with de, and also through the intersection of 
oa with ae. Hence, since these intersections are both known, 
oe can be drawn at once as shown in the figure ; and then 
OE can be drawn parallel to oe. This fixes the point E ; 
and DE and EA represent, in magnitude and direction, the 
required forces whose lines of action are de, ea. 

39. Problems in Equilibrium. II. — A rigid body is in 
equilibrium under the action of a system of non-parallel forces, 
all known, except two ; of these, the line of action of one 
and the point of application of the other are given. It is 
required to completely determine the unknown forces. 

Let the system consist of five forces to be represented in 
the usual way ; in magnitude and direction by lines marked 
AB, BC, CD, DE, EA ; and in lines of action by ab, be, ed, 
de, ea. Of these lines let be, cd, de, ea (Fig. 11), be given, and 
let Z be a given point of ab. Also let BC, CD, and DE be 
known, while EA, AB are unknown. First, draw the force 
polygon as far as possible, giving B, C, D, E, four consecutive 
vertices of the force 
polygon. The side EA 
must be parallel to ea, 
but its length is un- 
known, hence the ver- 
tex A of the force 
polygon cannot be 
fixed. Next, draw the 
funicular polygon so 




Fig. 11 



22 GRAPHIC STATICS. 

far as possible from the given data. Choose the pole O, and 
draw the rays OB, OC, OD, OE. The remaining ray, OA, 
cannot yet be drawn. Now, in the funicular polygon, the 
sides ob, oc must intersect on be ; oc and od must intersect 
on cd\ od a.nd ^^ must intersect on de ; oe and oa must inter- 
sect on ea ; and oa and ob must intersect on ab. Since L 
is the only known point of the line ab, let this point be 
taken as the point of intersection of oa and ob. Now draw 
ob through L parallel to OB, and successively oc, od, oe, par- 
allel to OC, OD, OE. We may now draw oa, joining L with 
the point in which oe intersects ea. This completes the funic- 
ular polygon. The force polygon can now be completed as 
follows : From O draw a line parallel to oa and from E a line 
in the known direction of EA ; their intersection determines 
A. This determines both EA and AB, and the force polygon 
is completely known. The line of action ab may now be 
drawn through L parallel to AB, and the forces are completely 
determined. 

40. Problems in Equilibrium. III. — A rigid body is in 
equilibrium under the action of any number of forces, of 
which three are known only in line of action ; the remaining 
forces being completely known. It is required to determine 
the unknown forces. 

Let the given forces be six in number, their lines of action 
being represented by ab, be, cd, de, ef, fa (Fig. 12), and let AB, 
BC, and CD be known, while the remaining three forces are un- 
known in magnitude and direction. The known data make it 
possible to draw at once three sides of the force polygon, 
namely, AB, BC, CD; and the four rays OA, OB, OC, OD, 
from any assumed pole O. Also, four sides of the funicular 
polygon, oa, ob, oe, od, may be drawn at once. But oe and of 
are unknown ; also DE, EF, and FA in the force polygon. 

Since any side of the funicular polygon can be drawn through 
any chosen point, let the polygon be started by drawing oa 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



23 



through the intersection of ef dsx^ fa. We may then draw suc- 
cessively ob, oc, od. Now, of is unknown in direction ; but it 
is to be drawn through the intersection of oa and fa, hence, 
whatever its direction, it intersects ef in the same point (since 
oa was drawn through the intersection of ef and fa). Hence, 
the two vertices of the funicular polygon falling on ef and fa 
coincide at the intersection of these two lines. We may there- 




fore draw 06 through this point and also through the point 
already found by the intersection of od and de. Now draw a 
line from O parallel to oe, and from D a line parallel to de ; 
their intersection is E. Draw from E a line parallel to ef, and 
from A a line parallel to fa ; their intersection determines F. 
The force polygon is now completely drawn, and DE, EF, FA 
represent, in magnitude and direction, the required forces. The 
remaining ray OF and the corresponding string of may now be 
drawn, but are not needed. 

The construction might have been made equally well by 
choosing the intersection of de and ef as the starting point, 
since two vertices of the funicular polygon may be made to 
coincide at that point. Or, the point of intersection of de and 
af might be chosen ; but in that case, the order of the forces 
should be so changed as to make de and af consecutive. If 
this were done the figures should be relettered to agree with 



24 



GRAPHIC STATICS. 



the order in which the forces were taken. It may be noticed 
that the direction of the string first drawn may be chosen arbi- 
trarily and the pole so taken as to correspond to the direction 
chosen. This is important in the treatment of the following 
special case. 

Case of inaccessible points of intersection. — It may happen 
that the lines of action de, ef and fa have no point of intersec- 
tion within convenient limits. When this is the case, the 
method just given may still be applied, but involves the geo- 
metrical problem of drawing a line through an inaccessible point. 
For example, if ef and fa intersect beyond the limits of the 
drawing, as shown in Fig. 13, we may proceed as follows: 
■Choose some point of ab and from it draw a line through the 
point of intersection of ef and fa. (This can be done by a 
method to be explained presently.) Let this line be oa. From 
the known point A of the force polygon draw a line parallel to 





Fig. 13 



■oa and choose a pole upon it. Draw the rays OB, OC, OD ; 
then the corresponding strings in the order ob, oc, od. From 
the point in which od intersects de draw a line to the inaccessible 
point of intersection of ef and fa ; this will be oe. The force 
polygon can now be completed just as in the preceding case. 

A line may be drawn through the inaccessible point of inter- 
section of two given lines by the following method : Let AC, 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



25 



\E 



F D 



^J^ " Q 

P 

Fig. 14 



; which proves that AC, BG, and PQ meet in a point. 



BD (Fig. 14), be the given lines, and let it be required to draw 

a line through their point of intersection from some point P. 

Draw PA intersecting AC and 

BD, and from C, any point of 

AC, draw a line CQ parallel to 

PA. From E, a point of BD, 

draw EA, EP ; also draw CF 

parallel to AE, and FQ parallel 

to EP. Then PQ is the line 

required. For, by the similar 

triangles, it is easily shown that 

AB ^CG 

BP~ GQ 

Exception. — If the lines of action of the three unknown 
forces meet in a point (or are parallel), the problem is impos- 
sible of solution, unless it happens that the resultant of the 
known forces acts in a line through the point of intersection 
of those three lines of action (or parallel to them) ; in which 
case the problem is indeterminate. These cases will not be 
discussed here. 

Remark. — In the problems treated in this and the two pre- 
ceding articles, it will be noticed that the forces should be taken 
in such an order that those which are completely known are 
consecutive. Otherwise the number of unknown lines in the 
force and funicular polygons will be increased. [For an- 
other method of solving this problem, see Levy's " Statique 
Graphique."] 

41. Examples. — i. A rigid beam AB rests horizontally 
upon supports at A and B, and sustains loads as follows : 
Its own weight of 100 lbs. acting at its middle point ; a load 
of 50 lbs. at (7 ; a load of 60 lbs. at D ; and a load of 80 lbs. 
at E. The successive distances between A, C, D, E, and B 
are 4 ft., 6 ft., 7 ft., and 10 ft. Required the upward pressures 
on the beam at the supports. 



26 GRAPHIC STATICS. 

2. A rigid beam AB is hinged at A, and rests horizontally 
with the end B upon a smooth horizontal surface. The beam 
sustains loads as in Example i, and an additional force of 4a 
lbs. is applied at the middle point at an angle of 45° with the 
bar in an upward direction. Required the pressures upon the 
beam at A and B. [The pressure at A may have any direc- 
tion, while the pressure at B must be vertically upward, i.e., 
at right angles to the supporting surface. Hence, this is a 
particular case of Problem II.] 

3. Let the end B of the bar rest against a smooth surface 
making an angle of 30° with the horizontal ; the remaining 
data being as in Example 2. 

4. A rigid bar 2 ft. long weighs 10 lbs., its center of gravity 
being 8 inches from one end. The bar rests inside a smooth 
hemispherical bowl of 15 inches radius. What weight must 
be applied at the middle point in order that the bar may rest 
when making an angle of 15° with the horizontal.-* Also, 
what are the reactions at the ends ? 

5. A uniform bar 20 inches long, weighing 10 lbs., rests 
with one end against a smooth vertical wall and the other 
end overhanging a smooth peg 10 inches from the wall. A 
weight P is suspended from the end so that the bar is in 
equilibrium when making an angle of 30° with the horizontal. 
Find P, and the pressures exerted on the bar by the wall 
and peg. 

42. Special Methods. — Certain problems can be treated 
more simply by other methods than by the general method 
of constructing the force and funicular polygons. This is 
sometimes true of the following : 

Problem. — A rigid body is held in equilibrium by four 
forces acting in known lines, only one being known in magni- 
tude and direction. It is required to completely determine 
the three remaining forces. (See Fig. 15.) 

Let the four forces have lines of action ab, be, cd, da, and let 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 27 

AB be drawn representing the magnitude and direction of the 
known force. Now the resultant of the forces whose lines 
of action are da and ad must act in a line passing through 
M, the point of intersection of these lines ; and the resultant 
of the other two forces must act in a line passing through 
JV, the point of intersection of dc and cd. But the two 



resultants must be equal and opposite and have the same 
line of action, else there could not be equilibrium. (Art. 36.) 
Hence, each must act in the line MJV. Draw BD parallel 
to MJV, and AD parallel to ad; the point D being determined 
by their intersection. Then DA represents in magnitude and 
direction the force acting in da, and DB the resultant of 
DA and AB. But BD must represent the resultant of the 
two remaining forces ; hence these two forces are represented 
by BC and CD drawn from B and D parallel respectively to 
dc and cd. 

This problem is a special case of that treated in Art. 40. 
But the construction here given will in many cases be the 
simpler one. 

Example. — A rigid body has the form of a square ABCD, 
the side AB being horizontal, and BC vertical. The weight 
of the body is 100 lbs., its center of gravity being at the inter- 
section of the diagonals. It is held in equilibrium by three 
forces as follows : Pj acting at C in the line AC ; P2 acting at 
D in the line AD ; and P3 applied at B and acting in the line 
joining B with the middle point of AD. Required to com- 
pletely determine Pi, P^, and P3. 



28 



GRAPHIC STATICS. 



§ 3. Resolution into Non-concitrrent Systems. 

43. To Replace a Force by Two Non-concurrent Forces. — 

This may be done in an infinite number of ways. The Hnes 
of action of the two components must intersect at a point on 
the line of action of the given force, and they must further 
satisfy the same conditions as concurrent forces. (Art. 21.) 

44. To Replace a Force by More Than Two Non-concurrent 
Forces. — This may be done by first resolving the given force 
into two forces by the preceding article, and then resolving 
one or both of its components in the same way. This problem 
and that of Art. 43 are indeterminate. (See note, Art. 20.) 
To make such a problem determinate, something must be 
specified regarding the magnitudes and lines of action of the 
required components. We shall consider some of the particular 
cases which are of frequent use in the treatment of practical 
problems. 

45. Resolution of a Force into Two Parallel Components. — 

Let it be required to resolve a force into two components 
having given lines of action parallel to its own. 

If the given force be reversed in direction, it will form with 
the required components a system in equilibrium. The com- 
ponents may then be determined by the method of Art. 38. 

Example. — Let the student solve two particular cases of 
this problem, taking the line of action of the given force (i) 
between those of the components, (2) outside those of the 
components. 

46. Resolution of a Force into Three Components. — Problem. 
■ — To resolve a force into three components having known lines 
of action. 

If the given force be reversed in direction, it will form, with 
the required forces, a system in equilibrium. Hence these 
forces may be determined by either of the two methods given 



RESOLUTION INTO NON-CONCURRENT SYSTEMS. 



29 



in Arts. 40 and 42. Or, the following reasoning may be 
employed, leading to the same construction as that of Art. 42. 

Let AD (Fig. 16) represent the given force in magnitude 
and direction, and ad its line of action ; the lines of action of 
the components being given as ad, be, cd. Since the given 
force may be assumed to act at any point in the line ad, let its 
point of application be taken at M, the point of intersection of 





Fig. 16 



ad and cd. Resolve it into two components acting in the lines 
cd and MN, N being the point of intersection of ab and be. 
These components are represented in magnitude and direction 
by AC, CD. Let AC, acting in the line J/yV (also marked ac), 
be resolved into components having lines of action ab, be. 
These components are given in magnitude and direction by 
AB, BC, drawn parallel respectively to ab, be. Hence, the 
given force, acting in ad, is equivalent to the three forces 
represented in magnitude and direction by AB, BC, CD, acting 
in the lines ab, be, cd. 

If the line of action of the given force does not intersect any 
one of the given lines ab, be, cd, within the limits of the draw- 
ing, it may be replaced by two components ; then each may be 
resolved in accordance with the above method, and the results 
combined. If the three lines ab, be, cd, are all parallel to ad, 
or if the four lines intersect in a point, the problem is inde- 
terminate. This is evident from the preceding articles, since, 
by methods already given, a part oi AD can be replaced by 
two components acting in any two of the given lines, as ab, 
be ; and the remaining part by two components acting in ab, cd; 



30 



GRAPHIC STATICS. 



or in be, cd. This construction evidently admits of infinite 
variation. 

For another method of solving the above problem, see Clarke's 
"Graphic Statics," p. i6. 

§ 4. Moments of Forees and of Co7tples. 

47, Moment of a Force. — Definition. — The moment of a 
force with respect to a point is the product of the magnitude of 
the force into the perpendicular distance of its line of action 
from the given point. The moment of a force with respect to 
an axis perpendicular to the force is the product of the magni- 
tude of the force by the perpendicular distance from the axis 
to the line of action of the force. 

If the moment is taken with respect to a point, that point is 
called the origin of moments. The perpendicular distance from 
the origin, or axis, to the line of action of the force is called 
the arm. 

Rotation tendency of a force. — The moment of a force 
measures its tendency to produce rotation about the origin, or 
axis. Thus, if a rigid body is fixed at a point, but free to turn 
about that point in a given plane, any force acting upon it in 
that plane will tend to cause it to rotate about the fixed point. 
The amount of this tendency will be proportional both to the 
magnitude of the force and to the distance of its line of action 
from the given point ; that is, to the moment of the force with 
respect to the point, as above defined. 

Rotation in any plane may have either of two opposite 
directions, which may be distinguished from each other by signs 
plus and minus. Rotation with the hands of a watch supposed 
placed face upward in the plane of the paper will be called 
negative, and the opposite kind positive. It would be equally 
legitimate to adopt the opposite convention, but the method 
here adopted agrees with the usage of the majority of writers. 

The sign of the moment of a force is regarded as the same 
as that of the rotation it tends to produce about the origin. 



MOMENTS OF FORCES AND OF COUPLES. 



31 



Moment represented by the area of a triangle. — If a triangle 
be constructed having for its vertex the origin of moments, and 
for its base a length in the line of action of a force, numerically 
equal to its magnitude, then the moment of the force is numeri- 
cally equal to double the area of the triangle. This follows at 
once from the definition of moment. 

48. Moment of Resultant of Two Non-parallel Forces. — Propo- 
sition. — The moment of the resultant of two non-parallel forces 
with reference to a point in their plane is equal to the algebraic 
sum of their separate moments with reference to the same point. 

In Fig. 17, let AB, BC, and AC represent in magnitude and 
direction two forces and their resultant ; and let ab, be, ac be 
their lines of action, intersecting in a jj. ^ , 

point N. Choose any point M as the \'"'\^"^r""^^7^ 

•origin of moments. Lay off NP = AB; \ J\^^— -— 1^^.^^ 

NQ^BC; and NR = AC. Then the o'V-'T^-r'^^^/ 
moments of the three forces are re- m'^/'^^W/ 

spectively equal to double the areas of ^-^\^^- — '"P ^q 
the triangles MNP, MNQ, MNR. ' ^ '' /^^/ 

These three triangles have a common ^^ / 

side MN, which may be considered the A e 

base ; hence their areas are proper- ■^^^" ^'^ 

tional to their altitudes measured from that side. These alti- 
tudes are PP\ QQ' , RR' , perpendicular to MN. Now, if PP" 
is parallel to MN, P"R' is equal to PP' ; also RP" equals 
QQ' (since they are homologous sides of the equal triangles 
NQQ', PRP"). Hence RR' = PP' + QQ', and therefore 

Area J/iW? = area MNP + 2lx&^. MNQ ; 

which proves the proposition. 

If the origin of moments is so taken that the moments of 
AB and BC have opposite signs, the demonstration needs modi- 
fication. The student should attempt the proof of this case for 
himself. 



z^ 



GRAPHIC STATICS. 



49. Moment of a Couple. — Definition. — The moment of a 
couple about any point in its plane taken as an origin is the 
algebraic sum of the moments of the two forces composing it 
with reference to the same origin. 

Proposition. — The moment of a couple is the same for every 
origin in its plane, and is numerically equal to the product of 
the magnitude of either force into the arm of the couple. 
(Art. 7.) 

The proof of this proposition can readily be supplied by the 
student. 

50. Moment of the Resultant of Any System. — Proposition. 
— The algebraic sum of the moments of any given complanar 
forces, with reference to any origin in their plane, is equal to 
the moment of their resultant force or resultant couple with 
reference to the same origin. 

The construction employed in proving this proposition is 
similar to that used in Art. 27, which the student may profit- 
ably review at this point. Referring to Fig. 9, let the given 
forces be represented in magnitude and direction by AB, BC, 
CD, DE, and in lines of action by ab, be, cd, de. Let the 
origin of moments be any point in the space diagram. As in 
Art. 27, replace AB by AO, OB, acting in lines ao, ob ; replace 
BC hy BO, OC, acting in lines bo, oc ; replace CD by CO, OD, 
acting in lines eo, od; and replace DE by DO, OE, acting in 
lines do, oe. (For brevity, we refer to a force as AB, meaning 
"the force represented in magnitude and direction by AB.") 

Now by Art. 48, we have, whatever the origin, 

Moment of /ii? = moment of y^C' + nioment of OB; 
" BC= " " B0+ " " OC; 
" CD= " " C0+ " " OD; 

" DE= " " D0+ " " OE. 

Since the forces represented by BO and OB have the same 
line of action, their moments are numerically equal but have 



MOMENTS OF FORCES AND OF COUPLES. 



33: 



opposite signs ; and similar statements are true of CO and 
OC, and of DO and OD. Hence, the addition of the above 
four equations shows that the sum of the moments of AB, 
BC, CD, DE, is equal to the sum of the moments of AO 
and OE. Now the given system has either a resultant force 
or a resultant couple. In the first case the resultant of the 
system is the resultant oi AO and OE, and its moment is 
equal to the algebraic sum of their moments, by Art. 48. 
In the second case (which occurs only when E coincides with. 
A), the resultant couple is composed oi AO and OE (which 
in this case are equal and opposite forces), and the moment 
of the couple is, by definition, equal to the algebraic sum of 
the moments oi AO and OE. Hence, in either case, the 
proposition is true. 

It should be noticed that the proof here given applies to 
systems of parallel forces, as well as to non-parallel systems. 
The proposition of Art. 48 could be extended to the case of 
any number of forces, by considering first the resultant of 
two forces, then combining this resultant with the third force, 
and so on ; but the method would fail if, in the process, it 
became necessary to combine parallel forces. The method 
here adopted is not subject to this failure. 

51. Condition of Equilibrium. — It follows from what has 
preceded, that if a given system is in eqnilibrinm, the alge- 
braic S2nn of the moments viust be zero, whatever the origin.. 
For, in case of equilibrium, AO and OE (Fig. 9) are equal and 
opposite and have the same line of action ; hence, the sum 
of their moments (which is the same as the sum of the 
moments of the given forces) is equal to zero. Conversely, 
If the algebraic snm of the moments is zero for every origin, 
the system must be in equilibrinm. For, if it is not, there 
is either a resultant force or a resultant couple. But the 
moment of a force is not zero for any origin not on its line 
of action ; and the moment of a couple is not zero for any 



34 



GRAPHIC STATICS. 



origin. For a fuller discussion of the conditions of equilibrium, 
see Arts. 58 and 59. 

52. Equivalent Couples. — Proposition. — If a system has for 
its resultant a couple, it is equivalent to any couple whose 
moment is equal to the sum of the moments of the forces 
of the system. 

For, as already seen (Art. 31), when the resultant is a 
couple, the force polygon is closed. Let the initial and final 
points of the force polygon coincide at some point A, and 
let O be the pole. Then the forces of the resultant couple 
are represented in magnitude and direction by AO, OA. Since 
the position of O is arbitrary, the force AO (or OA) may be 
made anything whatever in magnitude and direction. Also 
the line of action of the force AO may be taken so as to pass 
through any chosen point. Hence, the resultant couple may 
have for one of its forces any force whatever in the plane of 
the given system ; and the other force will have such a line 
of action that the moment of the couple will be equal to the 
sum of the moments of the given forces. 

This reasoning is equally true if the given system is a 
couple. Hence, a couple is equivalent to any other couple 
having the same moment. In other words, all couples whose 
moments are equal are eqnivalejit ; and conversely, all equiva- 
lent couples have equal moments. 

[Note. — The construction above discussed fails if all forces of the given system 
are parallel to the direction chosen for the forces of the resultant couple. For then 
the force polygon is a straight line, and if the pole is chosen in that line, the 
strings of the funicular polygon are parallel to the lines of action of the forces, and 
the polygon cannot be drawn. But in this case, the system may first be reduced to a 
couple whose forces have some other direction, and this couple may be reduced to 
one whose forces have the direction first chosen. Hence, the proposition stated 
holds in all cases.] 

53. Moment of a System. — Definition. — The moment of a 
system of forces is the algebraic sum of the moments of the 
forces of the system. 



GRAPHIC DETERMINATION OF MOMENTS. 



35 



54. Moments of Equivalent Systems. — Proposition. — The 
moments of any two equivalent systems of complanar forces 
with respect to the same origin ar# equal. 

This follows immediately from the preceding articles. For, 
if the two systems are equivalent, each is equivalent to the 
same resultant force or resultant couple, and the moments of 
the two systems are therefore each equal to the moment of 
this resultant and hence to each other. 

§ 5. Graphic Determination of Moments. 

55. Proposition. — If, through any point in the space dia- 
gram a line be drawn parallel to a given force, the distance 
intercepted upon it by the two strings corresponding to that 
force, multiplied by the pole distance of the force, is equal 
to the moment of the force with respect to the given point. 

By the strings "corresponding to" a given force are meant 
the two strings which intersect at a point on its line of action. 

Let AB (Fig. 18) represent the magnitude and direction of 
a force whose line of action is ab, and let M be the origin 
of moments. Let O be the pole, 

and OK { — H) the pole distance a a\b 3^'' 

of the given force. Draw the \ ""^-^^ \ ^.^o" \ 

strings oa, ob, and through M k\ -_;/^0 \"--.^^ \ 

draw a line parallel to ab, inter- \/''' \ '^^""A^. 

secting oa and ob in P and Q. Fig. is ".m 

Then it is to be proved that the 

moment of the given force with respect to 31 is equal to 

HxPQ. 

Let h equal the perpendicular distance of M from ab. Then 
the required moment is AB x //. But since the similar triangles 
OAB and RPQ have bases AB, PQ, and altitudes H, h, respec- 
tively, it follows that 

^ = - ; hence, ABxh = PQxH, 
PQ h 

which proves the proposition. 



36 



GRAPHIC STATICS. 



It should be noticed that PQ represents a length, while IT 
represents 2^ force magnitude. Hence, the moment of the given 
force with respect to M is equal to the moment of a force H 
with an arm PQ. [It may, in fact, be shown that the given 
force is equivalent to an equal force acting in the line PQ 
(whose moment about M is therefore zero), and a couple with 
forces of magnitude H, and arm PQ. For AB acting in ab is 
equivalent to AO and OB acting in ao and od respectively. 
Also, AO acting in ao is equivalent to forces represented by 
AK and KO acting respectively in PQ and in a line through 
P at right angles to PQ ; and OB may be replaced by forces 
represented by OK and KB, the former acting in a line through 
Q perpendicular to PQ, and the latter in PQ. But AK and 
KB are equivalent to AB ; hence, the proposition is proved.] 

56. Moment of the Resultant of Several Forces. — The mo- 
ment of the resultant of any number of consecutive forces in 
the force and funicular polygons may be found by a method 
similar to that just described. Thus, let Fig. 19 represent the 

force polygon and the 
funicular polygon for six 
forces, and let it be re- 
quired to find the moment 
of the resultant of the four 
forces represented in the 
force polygon by BC, CD, 
DE, and EF, with respect 
to any point M. The re- 




sultant of the four forces 
is represented by BE, and 
acts in a line through the intersection of ob and of. Through 
M draw a line parallel to BE, intersecting ob and of m P and Q 
respectively. Then PQ multiplied by OK, the pole distance of 
BE, gives the required moment. This method does not apply 
to the determination of the moment of the resultant of several 
forces not consecutive in the force polygon. 



SUMMARY OF CONDITIONS OF EQUILIBRIUM. 37 

57. Moments of Parallel Forces. — The method of Arts. 55 
and 56 is especially useful when it is desired to find the mo- 
ments of any or all of a system of parallel forces ; since with 
:such a system the pole distance is the same for all forces, and 
the moments are therefore proportional to the intercepts found 
iDy the above method. 

Example. — Assume five parallel forces at random ; choose 
an origin, and determine their separate moments, also the 
moment of their resultant, by the method of Arts. 55 and 56. 

§ 6. Siwiinary of Conditions of Eqiiilibrinm. 

58. Graphical and Analytical Conditions of Equilibrium Com- 
pared. — It has been shown (Art. 35) that the conditions of 
equilibrium for a system of complanar forces acting on a rigid 
l3ody are two in number : 

(1) The force polygon must close. 

{II) Any funicular polygon must close. 
The analytical conditions * are the following : 
(i) The algebraic sum of the resolved parts of the forces 
in any direction must be zero. 

(2) The algebraic sum of their moments for any origin must 
Tdc zero. 

The condition (i) is readily seen to be equivalent to (I). 
For if the sides of the force polygon be orthographically 
projected upon any line, their projections will represent in 
magnitude and direction the resolved parts of the several 
forces parallel to the line; and, if the force polygon is closed, 
the algebraic sum of these projections is zero, whatever the 
■direction of the assumed line. (See Art. 22.) It may also 
be seen that condition (II) carries with it (2). For, if every 
funicular polygon closes, the system is equivalent to two equal 
and opposite forces having the same line of action (Arts. 31, 

* See Minchitis Statics, Vol. I, p. 114. 



38 



GRAPHIC STATICS. 



35, and 50) ; and the sum of the moments of these two forces 
must be zero. 

A further comparison may be made. The analytical condi- 
tion (2) carries (i) with it ; and similarly the graphical condi- 
tion (II) carries with it the condition (I). That (2) includes 
(i) may be seen as follows : If the sum of the moments is zero 
for one origin M^, there can be no resultant couple, neither can 
there be a resultant force unless with a line of action passing 
through 3f■^^. If the sum of the moments is zero for a 
second origin M^, the resultant force, if one exists, must 
act in the line M1M2. If the sum of the moments is zero 
also for a third origin Mz, not on the line M1M2, there can 
be no resultant force. It follows at once that condition (i) 
must hold. 

That (II) includes (I) may be shown as follows : Let A and 
E be the first and last points of the force polygon, and choose 
a pole Oi. Then, if the funicular polygon closes, O^A and 
OiE are parallel. Choose a second pole O^, and, if the funic- 
ular polygon again closes, O^A and O2E are parallel. AE 
must then be parallel to O1O2, unless A and E coincide. 
Now, choose a third pole Oz, not on O1O2 ; if the funicular 
polygon for this pole closes, O^A and O^E must be parallel. 
But this is impossible unless A and E coincide ; that is, unless 
condition (I) holds. 

The last result may be reached in another way. With any 
pole O draw a funicular polygon and suppose it to close. The 
system is thus reduced to two forces acting in the same line 
oa. Hence, there is no resultant couple, and if there is a 
resultant force, its line of action is oa. With the same pole 
draw a second funicular polygon, the first side being <?'«', 
parallel to oa. If this polygon closes, there can be no result- 
ant force, for if one existed it would act in the line o'a' ; and 
there would thus be two resultant forces, acting in different 
lines oa and o'a\ which is impossible. 



SUMMARY OF CONDITIONS OF EQUILIBRIUM. 30 

59. Summary. — It is now evident that the conditions neces- 
sary to insiLve equilibrium may be stated in several different 
ways, both analytically and graphically. To summarize : 

A. Analytically : There will be equilibrium if either of the 
following conditions is satisfied : 

(i) The sum of the moments is zero for each of three points 
not in the same line. 

(2) The sum of the moments is zero for each of two points, 
and the sum of the resolved parts is zero in a direction not 
perpendicular to the line joining those two points. 

(3) The sum of the moments is zero for one point, and the 
sum of the resolved parts is zero for each of two directions. 

B. Graphically : There will be equilibrium if either of these 
three conditions is satisfied : 

(i) A funicular polygon closes for each of three poles not in 
the same line. 

(2) Two funicular polygons close for the same pole. 

(3) One funicular polygon closes and the force polygon closes. 



CHAPTER III. INTERNAL FORCES AND 
STRESSES. 

§ I. External and Internal Forces. 

60. Definitions. — It was stated in Art. 3 that every force 
acting upon any body is exerted by some other body. In what 
precedes, we have been concerned only with the effects pro- 
duced by forces upon the bodies to which they are applied. 
It has therefore not been needful to consider the bodies which 
exert the forces. It is now necessary to consider forces in 
another aspect. 

The forces applied to any particle of a body may be either 
external or internal. 

An external force is one exerted upon the body in question 
by some other body. 

An internal force is one exerted upon one portion of the body 
by another portion of the same body. 

It is important to note, however, that the same force may be 
internal from one point of view, and external from another. 
Thus, if a given body be conceived as made up of two parts, 
X and F, a force exerted tipon X by F is internal as regards 
the whole body, but external as regards the part X. Thus, let 

AB (Fig. 20) represent 
•< \ \ |> ^ a bar, acted upon by two 

A C Br 

forces of equal magnitude 
jpig'. so 

applied at the ends paral- 
lel to the length of the bar, in such a way as to tend to pull it 
apart. These two forces are exerted upon AB by some other 

40 



EXTERNAL AND INTERNAL FORCES. 41 

bodies not specified. If the whole bar be considered, the 
external forces acting upon it are simply the two forces 
named. 

But suppose the body under consideration is AC, a portion of 
AB. The external forces acting upon this body are (i) a force 
at A, already mentioned, and (2) a force at C, exerted upon AC 
by CB. This latter force is internal to the bar AB, but external 
to AC 

61. Conditions of Equilibrium Apply to External Forces. — In 

applying the conditions of equilibrium deduced in previous 
articles, it must be remembered that only external forces are 
referred to. It is also important to notice that the principles 
apply to any body or any portion of a body in equilibrium ; and 
the system of forces in every case must include all forces that 
are external to the body or portion of a body in question. 

Thus, if the bar AB (Fig. 20) be in equilibrium under the 
action of two opposite forces P and Q, applied at A and B 
respectively, as shown in the figure, the principles of equilibrium 
may be applied either to the whole bar, or to any part of it, as 
AC 

{a) For the equilibrium of the whole bar AB, we must have 
F=Q, these being supposed the only external forces acting on 
the bar. 

{b) For the equilibrium of AC, the force exerted upon AC hy 
CB must be equal and opposite to P. This latter force is 
external \.o AC, though internal to the whole bar. 

The method just illustrated is of frequent use in the investi- 
gation of engineering structures. It is often desired to deter- 
mine the internal forces acting in the members of a structure, 
and the general method is this : Direct the attention to such a 
portion of the whole structure or body considered that the 
internal forces which it is desired to determine shall be external 
to the portion in question. (See Art. 6^.) 



42 



GRAPHIC STATICS. 



2. External and Internal Stresses. 



62. Newton's Third Law. — Let X and Y be any two portions 
of matter ; then if X acts upon Y with a certain force, Y acts 
upon X with a force of equal magnitude in the opposite direc- 
tion. This is the principle stated in Newton's third law of 
motion, — that "to every action there is an equal and contrary 
reaction." It is justified by universal experience. 

63. Stress. — Definition. — Two forces exerted by two por- 
tions of matter upon each other in such a way as to constitute 
an action and its reaction, make up a stress. 

Ilhi.strations. — The earth attracts the moon with a certain 
force, and the moon attracts the earth with an equal and opposite 
force. The two forces constitute a stress. 

Two electrified bodies attract (or repel) each other with equal 
and opposite forces. These two forces constitute a stress. 

Any two bodies in contact exert upon each other equal and 
opposite pressures (forces), constituting a stress. 

By the magnitude of a stress is meant the magnitude of either 
of its forces. 

64. External and Internal Stresses. — It has been seen that 
two portions of matter are concerned in every stress. Now the 
two portions may be regarded either as separate bodies, or as 
parts of a body or system of bodies which includes both. 

A stress acting between two parts of the same body (or 
system of bodies) is an internal stress as regards that body 
or system. 

A stress acting between two distinct bodies is an external 
stress as regards either body. 

It is important to notice that the same stress may be internal 
from one point of view, and external from another. Thus, if 
a given body be considered as made up of two parts X and Y, 
a stress exerted between X and Y is internal to the whole body, 
but external to either X or Y. 



EXTERNAL AND INTERNAL STRESSES. 



43 




Illustration. — Consider a body AB (Fig. 21) resting upon a 
second body Y, and supporting another body X, as shown. If 
the weight of the body AB be disregarded, 
the forces acting upon it are (i) the down- 
ward pressure (say P) exerted by X at the 
surface A, and the upward pressure (say Q) 
exerted by Fat ^ ; these forces being equal 
and opposite, since the body is in equiHbrium. 
Now the body X is acted upon by AB with 
a force equal and opposite to P, and these -^^^- ^^ 

two forces constitute a stress which is external to AB. There 
is also an external stress exerted between AB and Y at B. 
But let AB be considered as made up of two parts, AC and CB. 
Then (Art. 60) CB exerts upon AC 2, force upward, and AC 
exerts upon CB a force downward. These two forces are an 
action and its reaction, and constitute a stress which is internal 
to the body AB. This same stress is, however, external to 
either AC or CB. An equivalent stress evidently exists at 
every section between A and B. (When we refer to the force 
acting upon CB at C, we mean the resultant of all forces exerted 
upon the particles of CB by the particles of AC This resultant 
is made up of very many forces acting between the particles. 
Also the stress at C means the stress made up of the two 
resultants of the forces exerted by AC a.nd CB upon each other.) 

65. Three Kinds of Internal Stress. — It is evident that the 
internal stress at C in the body AB (Fig. 20) depends upon 
the external forces applied to the body. If the forces at A and 
B cease to act, the forces exerted by AC and CB upon each 
other become zero. If the forces at A and B are reversed in 
direction, so also are those at C (As a matter of fact, the 
particles of AC exert forces upon those of CB, even if the 
external forces do not act. But if the external forces applied 
to CB are balanced, the resultant of the forces exerted on CB 
by ACis zero.) 



44 



GRAPHIC STATICS. 



The nature of the internal stress at any point in a body is 
thus seen to depend upon the external forces applied to the body. 

Now, if we consider two adjacent portions of a body (as 
the parts X and Y, Fig. 22) separated by a plane surface, the 

external forces may have either of three 

-; ( [ M tendencies : (i) to pull X and Y apart 

in a direction perpendicular to the plane 
^^"" of separation; (2) to push them together 

in a similar direction ; (3) to slide each over the other along the 
plane of separation. Corresponding to these three tendencies, 
the stress between X and Y may be of either of three kinds : 
tensile, compressive, or sheaving. 

A tensile stress is such as comes into action to resist a 
tendency of the two portions of the body to be pulled apart in 
the direction of the normal to their surface of separation. 

A compressive stress is such as comes into action to resist a 
tendency of X and Y to move toward each other along the 
normal to the surface. 

A shearing stress is such as acts to resist a tendency of X 
and Fto slide over each other along the surface between them. 

In case of a tensile stress, the force exerted by X upon Y 
has the direction from Y toward X; and the force exerted by 
Y upon X has the direction from X toward Y. 

In case of a compressive stress, the force exerted by X upon 
Fhas the direction from Jf toward Y; and the force exerted by 
Fupon Jf has the direction from F toward X. 

In the case of a shearing stress, the force exerted by X 
upon Fmay have any direction in the plane of separation; the 
force exerted by F upon X having the opposite direction. 

If X and F are separate bodies, instead of parts of one 
body, a similar classification may be made of the kinds of 
stress between them ; but with these we shall have no occasion 
to deal. The terms tensile stress, compressive stress, and shear- 
ing stress (or tension, compression, and shear) are usually applied 
only to internal stresses. 



EXTERNAL AND INTERNAL STRESSES. 



45 



66. Strain. — In what has preceded, the bodies dealt with 
have been regarded as rigid ; that is, the relative positions of 
the particles of any body have been regarded as remaining 
unchanged. But, as remarked heretofore, no known body is 
perfectly rigid. If no external forces act upon a body, its 
particles take certain positions relative to each other, and the 
body has what is called its natural shape and size. If external 
forces are applied, the shape and size will generally be changed ; 
the body is then said to be in a state of strain. The deforma- 
tion produced by any system of applied forces is called the 
strain due to those forces. The nature of this strain in any 
case depends upon the way in which the forces are applied. It 
is unnecessary to treat this subject further at this point, since 
we shall at present be concerned only with problems in the 
treatment of which it will be sufficiently correct to regard the 
bodies as rigid. 

[Note. — There is a lack of uniformity among writers in regard to the meanings 
attached to the words stress and strain. It may, therefore, be well to explain again 
at this point the way in which these words are used in the following pages. The 
word stress should be employed only in the sense above defined, as consisting of two 
equal and opposite forces constituting an action and its reaction. The two forces are 
exerted respectively by two bodies or portions of matter upott each other. An 
internal stress is a stress between two parts of the same body. An internal force is 
one of the forces of an internal stress. It is intended in what follows to use the 
word? " internal stress " (or simply " stress ") only when both the constituent forces 
are referred to; and when only one of the forces is meant, to use the words "inter- 
nal force" (or simply "force"). It will be noticed, therefore, that in the following 
pages the words '■'■force in a bar " are frequently used where many writers would say 
" stress.'" This departure from the usage of many high authorities seems justified by 
the following considerations : (i) It agrees with the usage which is being adopted 
by the highest authorities in pure mechanics. (2) It is desirable that the nomencla- 
ture of technical mechanics shall agree with that of pure mechanics, so far as they 
deal with the same conceptions. The definition of strain above given is in conform- 
ity with the usage of the majority of the more recent text-books. But it is not rare 
to find in technical literature the word strain used in the sense of internal stress as 
above defined. Such use of the word should be avoided.] 



46 GRAPHIC STATICS. 

§ 3. Determination of Internal Stresses. 

67. General Method. — The stresses exerted between the 
parts of a body may or may not be completely determinate by 
means of the principles already deduced. But in all cases 
these principles suffice for their partial determination. The 
general method employed is always the same, and will now be 
illustrated. As heretofore we deal only with complanar forces. 
Let XY (Fig. 23) represent a body in equilibrium under the 
action of any known external forces as shown. Now conceive 
the body to be divided into two parts as X and Y, separated 
by any surface. The particles of X near the surface exert 
upon those of Y certain forces, and are in return acted 

upon by forces exerted by 
the particles of Y. These 
forces are internal as re- 
gards the whole body. In 
order to determine them so 
far as possible, we proceed 
as follows : Let the resultant of all the forces exerted by Y 
upon X be called T\ then 2" is either a single force or a couple. 
Now apply the conditions of equilibrium to the body X. The 
external forces acting on X are Pi, P^, Pz, and T. Since P^, P^, 
and 7^3 are supposed known, T can be determined. In fact, 
T is equal and opposite to the resultant of Pj, P^, and P^. 

So much can always be determined. But T is the resultant 
of a great number of forces acting on the various particles 
of X; and these separate forces cannot in general be deter- 
mined by methods which lie within the scope of this work. 

The general principle just illustrated may be stated as 
follows : 

If a body in equilibrium under any external forces be conceived 
as made up of two pa^'ts X and Y, then the internal forces 
exerted by X upon Y, together with tJie external forces acting on 
Y, form a system in equilibrium. 




DETERMINATION OF INTERNAL STRESSES. 



47 



As an immediate consequence, we may state that the result- 
ant of the forces exerted by X upon V is eqnivalent to the residt- 
ant of the external forxes acting on X ; and is equal and opposite 
to the resultant of the external forces acting upon Y. 

Example. — Assume a bar of known dimensions, and the 
magnitudes, directions, and points of application of five forces 
acting on it. Then (i) determine a sixth force which will 
produce equilibrium ; and (2) assume the bar divided into two 
parts and find the resultant of the forces exerted by each part 
on the other. 

6Z. Jointed Frame. — In certain ideal cases (corresponding 
more or less closely to actual cases), the internal forces may be 
more completely determined. The most important of these 
cases is that which will be now considered. Conceive a rigid 
body made up of straight rigid bars hinged together at the ends. 
Assume the following conditions : 

(i) The hinges are without friction. 

(2) All external forces acting on the body are applied at 
points where the bars are joined together. 

The meaning of these conditions will be seen by reference to 
Fig. 24. The three bars X, Y, and Z are connected by a " pin 
joint," the end of each bar having a hole 
or "eye" into which is fitted a pin. 
(Of course the three bars cannot be in 
the same plane, but they may be nearly 
so, and will be so assumed in what fol- 
lows.) Condition (i) is satisfied if the 
pin is assumed frictionless. The effect of this is that the force 
exerted upon the pin by any bar (and the equal and opposite 
reaction exerted upon the bar by the pin) act in the normal 
to the surfaces of these bodies at the point of contact ; and, 
therefore, through the centers of the pin and the hole. Condi- 
tion (2) means that any external force (that is, external to the 




48 



GRAPHIC STATICS. 



whole body) applied to any bar is applied to the end and in a 
line through the center of the pin. 

With the connection as shown, the bars do not exert forces 
upon each other directly. But each exerts a force upon the 
pin, and any force exerted by F or Z upon the pin causes an 
equal force to be exerted upon X. (This is seen by applying- 
the condition of equilibrium to the pin.) Hence, in considering- 
the forces acting upon any bar as X, we may disregard the pin 
and assume that each of the other bars acts directly upon X. 
By what has been said, all such forces exerted upon X by the 
other bars meeting it at the joint may be regarded as acting at 
the same point — the center of the pin. We therefore treat the 
bars as mere "material lines," and regard all forces exerted on 
any bar (whether by the other bars or by outside bodies) as 
applied at the ends of this "material line." 

Since, with these assumptions, all forces acting on any bar 
MN (Fig. 25) are applied either at M or N, the forces applied 
at M must balance those applied at N. The resultants of the 
two sets must therefore be equal and opposite, and have the 
same line of action — namely MN. Further, it follows that 
the stress in the bar, acting across any plane perpendicular to 
its length, is a direct tension or compression. 



69. Internal Stresses in a Jointed Frame. — Let Fig. 25 rep- 
resent a jointed frame such as above described, in equilibrium 

under any known external forces. 
Let us apply the general method 
of Art. 6y to this case. Divide 
the body into two parts, X and F, 
by the surface AB as shown. Now 
apply the conditions of equilibrium 
to the body X. The system of 
forces acting upon this body consists of Pi, P-i, Pe, and the forces 
exerted by Fupon X \r\ the three members cut by the surface 
AB. By Art. 6^ the lines of action of these forces are known. 



ff 


S^r— 1— -. 


/p. 


i^ 


Hj 


xm 


Pi 


Fig,. s:S 





DETERMINATION OF INTERNAL STRESSES. 49 

being coincident with the axes of tlie members cut. Hence, 
the system in equiUbrium consists of six forces, three com- 
pletely known, and three known only in lines of action. 
The determination of the unknown forces in magnitude and 
direction is then a case under the general problem discussed in 
Art. 40. 

NatiLre of the stresses. — As soon as the direction of the 
force acting upon X in any one of the members cut is known, 
the nature of the stress in that member (whether tension or 
compression) is known. For a force tozvard X denotes com- 
pression ; while a force aivay from X denotes tension. (Art. 
65.) 

If a section can be taken cutting only two members, the 
forces in these may be found by the force polygon alone. The 
same is true, if any number of members are cut, but the stresses 
in all but two are known. 

The methods described in the last three articles will find 
frequent application in the chapters on roof and bridge trusses, 
Part II. 

Example. — Assume a jointed frame similar to the one shown 
in Fig. 25, and let external forces act at all the joints. Then 
(i) assume all but three of the forces known in magnitude and 
direction and determine the remaining three so as to produce 
equilibrium. (2) Take a section cutting three members and 
determine the stresses in those members. 

70. Indeterminate Cases. — If, in dividing the frame, more 
than three members are cut, the number of unknown forces is. 
too great to admit of the determination of their magnitudes. 
In such a case, it may happen that a section elsewhere through 
the body will cut but three members ; and that after the deter- 
mination of the stresses in these three, another section can be 
taken cutting but three members whose stresses are unknown. 
So long as this can be continued, the determination of the 



so 



GRAPHIC STATICS. 




internal stresses can proceed. Thus, in Fig. 26, if a section 
be first taken at AB, there are four unknown forces to be 
determined. But, if the section A'B' be first taken, the 
stresses in the three members cut may be determined ; after 

which the section AB will 
\A' introduce but three unknown 

stresses. 

There may, however, be 
cases in which the stresses 
cannot all be determined by 
any method. With such ac- 
tually indeterminate cases we shall not usually have to deal. It 
should be noticed, also, that even when only three members 
are cut, the problem is indeterminate (f tJiese tJiree intersect in 
.a point. As in the case just discussed, this indeterminateness 
may be either actual or only apparent ; in the latter case it may 
be treated as above indicated. 

No attempt is here made to develop all methods that are 
applicable or useful in the determination of stresses in jointed 
frames. Some of these are best explained in connection with 
the actual problems giving rise to them. We have sought here 
only to explain and clearly illustrate general principles. 

71. Funicular Polygon Considered as Jointed Frame. — Let 

ab, be, cd, da (Fig. 27) be the lines of action of four forces in 
equihbrium, the force polygon being ABCDA. Choosing a 




S7 




pole, draw any funicular polygon, as the one shown. Now let 
the body upon which the forces act be replaced by a jointed 



DETERMINATION OF INTERNAL STRESSES. 



51 



frame whose bars coincide with the sides of the funicular 
polygon. If at the joints of this frame the given forces be 
.applied, the frame will be in equilibrium ; and each bar will 
sustain a tension or compression whose magnitude is repre- 
sented by the corresponding ray of the force diagram. 

To prove this, we apply the "general method" of Art. 6j. 
Consider any joint (as the intersection of oa and ob), and let 
the frame be divided by a plane cutting these two members. 
Then the portion of the frame about the joint is acted upon 
by three forces : AB, acting in the line ab, and forces acting in 
the bars cut, their lines of action being oa, ob. If the bar 
oa sustains a compression and ob a tension, their magnitudes 
being represented by OA and BO respectively, the portion 
of the frame about the joint will be in equilibrium. Hence, 
the tendency of the force AB is to produce the stresses men- 
tioned in the bars oa, ob. In the same way it may be shown 
that the tendency of the force BC is to produce in ob and 
■oc tensile stresses of magnitudes OB and CO, respectively. 
Applying the same reasoning to each joint, it is seen that every 
part of the frame will be in equilibrium if the bars sustain 
stresses as follows : The bar oa must sustain a compression 
OA ; ob a tension OB; oc a tension OC; and od ■&. compression 
OD. Hence, if the bars are able to sustain these stresses, the 
frame will be in equilibrium. 

If the stress in any member of the frame is a tension, that 
member may be replaced by a flexible string. This is the 
origin of the name string as applied to the sides of the funic- 
ular polygon. This name is retained for convenience, but, as 
just shown, it is not always appropriate. 

72. Outline of Subject. — The foregoing pages, embracing 
Part I, have been devoted to a development of the principles 
of pure statics. We pass next to the application of these 
principles to special classes of problems. 

Part II treats of the determination of internal stresses in 



52 



GRAPHIC STATICS. 



engineering structures. Only " simple " structures are consid- 
ered, — that is, those whose discussion does not involve the 
theory of elasticity. The structures considered include roof 
trusses, beams, and bridge trusses. 

Part III develops the graphic methods of determining cen- 
troids (centers of gravity) and moments of inertia of plane 
areas, including a short discussion of "inertia-curves." 



Part II. 
STJ^BSSBS IN SIMPLE STRUCTURES. 

CHAPTER IV. INTRODUCTORY. 

§ I. Outline of Principles and Methods. 

73. The Problem of Design. — When any structure is sub- 
jected to the action of external forces, there are brought into 
action certain internal stresses in the several parts of the struc- 
ture. The nature and magnitudes of these stresses depend upon 
the external forces acting (Art. 65). In designing the structure, 
each part must be so proportioned that the stresses induced in 
it will not become such that the material cannot sustain them 
without injury. 

To determine these internal stresses, when the external forces 
are wholly or partly given, is the problem of design, so far as 
it will be here treated. 

74. External Forces. — The external forces acting on a struc- 
ture must generally be completely known before the internal 
stresses can be determined. These external forces are usually 
only partly given, and the first thing necessary is to determine 
them fully. 

The external forces include (i) the loads which the structure 
is built to sustain, and (2) the reactions exerted by other bodies 
upon the structure at the points where it is supported. The 
former are known or assumed at. the outset and the latter are 
to be determined. 

53 



54 



GRAPHIC STATICS. 



Fig. 2 8 



75. Two Classes of Structures. — Structures may be divided 
into two classes, according as they may or may not be treated 
as rigid bodies in determining the reactions. This may be 
illustrated as follows : 

Let a bar AB (Fig. 28) be supported in a horizontal position 

at two points A and B, the supports being smooth so that the 

jP pressures on the bar at 

^ ^ ^ A and B are vertical. 

Let a known load P be 
applied to the beam at a 
given point, and let it be required to determine the reactions at 
A and B. 

This is a determinate problem ; for there are three parallel 
forces in equilibrium, two being known only in line of action. 
This problem was solved in Art. 38. 

But let AC (Fig. 29) be a rigid bar supported at three points^ 
A, B, and C, the reactions at those points being vertical. 

P B P 




I 



WA 




Let any known loads 

be applied at given 

points, and let it be 

required to determine 

Fig. 39 the three reactions. 

This problem is indeterminate ; for any number of sets of 

values of the three reactions may be found, which, with the 

applied loads, would produce equilibrium if acting on a rigid 

body. (See Art. 40.) 

Since, however, an actual bar is not a perfectly rigid body, 
such a problem as the one just stated is, in reality, determinate. 
But it cannot be solved without making use of the elastic 
properties of the material of which the body is composed. 

The two classes of problems are, therefore, the following : 
(i) those in which the reactions can be determined by treating 
the structure as a rigid body, and (2) those in which the 
determination of the reactions involves the theory of elasticity. 
We shall at present deal only with the former class of prob- 



OUTLINE OF PRINCIPLES AND METHODS. 



55 



lems. Structures coming under this class will be called simple 
structures. 

^6. Truss. — A truss is a structure made up of straight bars 
with ends joined together in such a manner that the whole acts 
as a single body. The ends of the bars are, in practice, joined 
in various ways ; but in determining the internal stresses, the 
connections are assumed to be such that no resistance is offered 
by a joint to the rotation of any member about it. Such a 
structure to be indeformable must be made up of triangular 
elements ; for more than three bars hinged together in the 
form of a polygon cannot constitute a rigid whole. If the- 
external forces are applied to the truss only at the points where 
the bars are joined, the internal stress at any section of a mem- 
ber will be a simple tension or compression, directed parallel to 
the length of the bar. (See Art. 68.) 

The most important classes of trusses are roof trusses and 
bridge trusses. The methods used in discussing these classes. 
are, of course, applicable to any framed structures under similar 
conditions. 

yj. Loads on a Truss. — The loads sustained by a truss may 
be either fixed or moving. A fixed (or dead) load is one whose 
point of application and direction remain constant. A moving 
(or live) load is one whose point of application passes through 
a series of positions. Fixed loads may be either permane^tt or 
temporary. 

The loads on a roof truss are usually all fixed, but are of 
various kinds, viz., the weight of the truss itself and of the roof 
covering, which is a permanent load ; the weight of snow 
lodging on the roof, and the pressure of wind, both of which 
are temporary loads. 

A bridge truss supports both fixed and moving loads. The 
former include the weight of the truss itself, of the roadway, 
of all lateral and auxiliary bracing (permanent loads) ; and of 
snow (a temporary load). The latter consist of moving trains. 



56 GRAPHIC STATICS. 

in the case of railway bridges, and of teams or crowds of animals 
or people in the case of highway bridges. 

78. Combination of Stresses Due to Different Causes. — When 
a truss is subject to a variety of external loads, it is often 
convenient to consider the effect of a part of them separately. 
If tensile and compressive stresses are distinguished by signs 
plus and minus, the stress in any member due to the combined 
action of any number of loads is equal to the algebraic sum of 
the stresses due to the loads acting separately. 

A proof of this proposition might be given ; but it may be 
accepted as sufficiently evident without formal demonstration. 

79. Beams. — Another class of bodies to be treated is 
included under the name beam. 

A beam may be defined as a bar (usually straight) resting on 
supports and carrying loads. The loads and reactions are 
commonly applied in a direction transverse to the- length of the 
bar ; but this is not necessarily the case. 

The internal stresses in any section of a beam are less 
simple than those in the bars of an ideal jointed frame such as 
a truss is assumed to be. A discussion of beams is given in 
Chap. VI. 

80. Summary of Principles Needed. — It will be well to 
summarize at this point the main principles and methods which 
will be employed in the discussion of the problems that follow. 

The general problem presented by any structure consists 
of two parts : {a) the determination of the unknown external 
forces (or reactions) and {b) the determination of the internal 
stresses. 

{a) In the case of simple structures the unknown reactions 
are usually two in number, and the cases most commonly pre- 
sented are the following : 

1st. — Their lines of action are known and parallel. 

2nd. — The line of action of one and the point of application 
of the other are known. 



OUTLINE OF PRINCIPLES AND METHODS. 



57 



Since all the external forces form a system in equilibrium, 
these two cases fall under the general problems discussed in 
Arts. 38 and 39. 

{b) In the case of a jointed frame or truss, the lines of 
action of all internal forces are known, since they coincide 
with the axes of the truss members. (Art. 6d>.^ In determin- 
ing their magnitudes we may have to deal with the following 
problems in equilibrium : 

1st. — The system in equilibrium may be completely known, 
except the magnitudes of two forces. 

2nd. — The magnitudes of three forces may be unknown. 

The first case may be solved by simply making the force 
polygon close. (See Art. 35.) 

The second case may be solved by the method of Art. 40, 
which consists in making the force and funicular polygons 
close ; or by the method of Art. 42 ; or by the principle of 
moments (Art. 51). 

The student should be thoroughly familiar with the prob- 
lems and principles here referred to. In the following chapters 
we proceed to their application. 

81. Division of the Subject. — The subject of the design of 
structures, so far as here dealt with, will be treated in three 
divisions. The first relates to framed structures sustaining 
only stationary loads ; the second to beams sustaining both 
fixed and moving loads ; the third to framed structures sustain- 
ing both fixed and moving loads. Among structures of the 
first class, the most important are roof trusses ; hence, these 
are chiefly referred to in the next chapter. For a similar 
reason, the chapter devoted to the third class of structures 
refers principally to bridge trusses. The chapter on beams 
precedes that on bridge trusses, for the reason that the methods 
used in dealing with a beam under moving loads form a useful 
introduction to those employed in treating certain classes of 
truss problems. 



CHAPTER V. ROOF TRUSSES. —FRAMED STRUC- 
TURES SUSTAINING STATIONARY LOADS. 

§ I. Loads 071 Roof Trusses. 

82. Weights of Trusses. — Among the loads to be sustained 
by a roof truss is the weight of the truss itself. Before the 
structure is designed, its weight is unknown. But, since it is 
necessary to know the weight in order that the design may be 
correctly made, the method of procedure must be as follows : 

Make a preliminary estimate of the weight, basing it upon 
knowledge of similar structures ; or, in the absence of such 
knowledge, upon the best judgment available. Then design 
the various truss members, compute their weight, and com- 
pare the actual weight of the truss with the assumed weight. 
If the difference is so great as to materially affect the design of 
the truss members, a new estimate of weight must be made, 
and the computations repeated or revised. No more than one 
or two such trials will usually be needed. 

As a guide in making the preliminary estimate of weight, 
the following formulas may be used. They are taken from Mer- 
riman's " Roofs and Bridges," being intended to represent 
approximately the data for actual structures, as compiled by 
Ricker in his "Construction of Trussed Roofs." 

Let /=span in feet ; « = distance between adjacent trusses in 
feet ; W=- total weight of one truss in pounds. Then for 
wooden trusses 

and for wrought iron trusses 

^=1 «/(i+Jo/). 
58 



LOADS ON ROOF TRUSSES. 



59 



83. Weight of Roof Covering. — The weight of roof covering- 
can be correctly estimated beforehand from the known weights 
of the materials. The following data may be employed, in the 
absence of information as to the specific material to be used. 
(See Merriman's " Roofs and Bridges," p. 4.) The numbers 
denote the weight in pounds per square foot of roof surface. 

Shingling : Tin, i lb. ; wooden shingles, 2 to 3 lbs. ; iron, i 
to 3 lbs. ; slate, 10 lbs. ; tiles, 12 to 25 lbs. 

Sheathing : Boards i in. thick, 3 to 5 lbs. 

Rafters : 1.5 to 3 lbs. 

Purlins : Wood, i to 3 lbs. ; iron, 2 to 4 lbs. 

Total roof covering, from 5 to 35 lbs. per square foot of roof 
surface. 

84. Snow Loads. — The weight of snow that may have to be 
borne will differ in different localities. For different sections 
of the United States the following may be used as the maximum 
snow loads likely to come upon roofs. 

Maximum for northern United States, 30 lbs. per square foot 
of horizontal area covered. 

For latitude of New York or Chicago, 20 lbs. per square foot. 

For central latitudes in the United States, 10 lbs. per square foot. 

The above weights are given in Merriman's " Roofs and 
Bridges." They are in excess of those used by some Bridge 
and Roof companies. 

85. Wind Pressure Loads. — The intensity of wind pressure 
against any surface depends upon two elements : {a) the velocity 
of the wind, and {b) the angle between the surface and the 
direction of the wind. 

Theory indicates that the intensity of wind pressure upon a 
surface perpendicular to the direction of the wind should be 
proportional to the square of the velocity of the wind relative to 
the surface. As an approximate law this is borne out by 
experiment. If / denotes the pressure per unit area, and v 
the velocity of the wind, the law is expressed by the formula 



6o 



GRAPHIC STATICS. 



p — kv^. Here k is proportional to the density of air. Its 
numerical value may be taken as 0.0024, if the units of force, 
length, and time are the pound, foot, and second respectively. 

If the wind strikes a surface obliquely, experiment shows 
that the resulting pressure has a direction practically normal to 
the surface. The tangential component is inappreciable, owing 
to the very slight friction between air and any fairly smooth 
surface. The intensity of the normal pressure depends upon 
the angle at which the wind meets the surface. 

For a given velocity of wind let p^ denote the normal pressure 
per unit area, when the direction of wind makes an angle a 
with the surface, and /„ the pressure per unit area due to the 
same wind striking a surface perpendicularly. Then the follow- 
ing formula * has been given : 

2 sin a ^ 
po. = ■ o Pn. 

I + sm^ a 

It will rarely be necessary to use values of /„ greater than 
50 lbs. per square foot. The following table gives values of 
the coefficient of /„ in the above formula for different values 
of a. The value of /„ may be taken as from 40 to 50 lbs. per 
square foot. 



a 


2 sin a 
I + sin'2 a 


a 


2 sin a 
I + sin^o 


0° 

10° 

20° 
30° 
40° 


0.00 

0.34 
0.61 
0.80 
0.91 


50° 
60° 
70° 
80° 
90° 


0.97 
0.99 
1. 00 
1. 00 
1. 00 



* This formula is given by various writers. It is cited by Langley (" Experiments in 
Aerodynamics," p. 24), who attributes it to Duchemin. Professor Langley's elaborate 
experiments show so close an agreement with the formula that it may be used without hesi- 
tation in estimating the pressure on roofs. 



ROOF TRUSS WITH VERTICAL LOADS. 6l 

§ 2. Roof Truss zvith Vertical Loads. 

Z6. Notation. — The method of determining internal stresses 
in the case of vertical loading will be explained by reference to 
the form of truss shown in Fig. 30. The method will be seen 
to be independent of the particular form of the truss. 

For designating the truss members and the lines of action of 
external forces a notation will be employed similar to that used 
in previous chapters. Let each of the areas in the truss dia- 
gram be marked with a letter or other symbol as shown in 
Fig. 30 ; then the truss member or force-line separating any 
two areas may be designated by the two symbols belonging to 
those areas. Thus, the lines of action of the external forces 
are ab, be, ed, etc., and the truss members are gh, hb, hi, etc. 
It is to be noticed that the lines representing the truss mem- 
bers represent also the lines of action of forces, — namely, the 
internal forces in the members. The joint, or point at which 
several members meet, may be designated by naming all the 
surrounding letters. Thus, beih, Jiijg are two such points. 

87. Loads and Reactions. — The loads now considered are 
assumed to be applied in a vertical direction, and to act at the 
upper joints of the truss. This assumption as to the points of 
application may in some cases represent very nearly the facts ; 
in other cases the loads will, in reality, be applied partly at 
intermediate points on the truss members. If the latter is the 
case, the load borne upon any member is assumed to be divided 
between the two joints at its ends. In this case the member 
will be subject not only to direct tension or compression, but to 
bending. With the latter we are not here concerned, although, 
it must always be considered in designing the member. 

The ends of the truss are supposed to be supported on hori- 
zontal surfaces, and the reaction at each point of support is 
assumed to have a vertical direction. 

If the loading is symmetrical with reference to a vertical line 



62 GRAPHIC STATICS. 

through the middle of the truss, it is evident that each reaction 
is equal to half the total load. If the loading is not symmetri- 
cal, the reactions cannot be determined so simply. They may, 
however, be readily computed by either graphic or algebraic 
methods. Graphically, the problem is identical with that solved 
in Art. 38. The truss is treated as a rigid body, the external 
forces acting upon it being the loads and reactions, which form 
a system of parallel forces in equilibrium. Two of these forces 
(the reactions) are unknown in magnitude, but known in line 
of action. The construction for determining their magnitudes 
is as follows : 

Draw the force polygon ABCDEF for the five loads ; choose 
a pole O, draw rays OA, OB, OC, etc., and draw the funicular 
polygon as shown in Fig. 30. The two polygons are to be 
completed by including the reactions FG, GA, and both poly- 
gons must close. We may draw first og, the closing line of the 
funicular polygon, and then the ray OG parallel to it, thus 
determining the point G in the force polygon. The reactions 
are now shown in magnitude and direction by FG and GA. 

88. Determination of Internal Stresses. — When the external 
forces are all known, the internal stresses may be found very 
readily. The only principle needed is, that for any system of 
forces in equilibrium, the force polygon must close. The con- 
struction will now be explained. 

Considering any joint of the truss (Fig. 30) as gJiijg, fix the 
attention upon the portion of the truss bounded by the broken 
line in the figure. This portion is a body in equilibrium under 
the action of four forces whose lines of action coincide with the 
axes of the four bars gh, hi, ij, jg, respectively. These forces 
are intemial as regards the truss as a whole, but extejmal to the 
part in question ; each force being one of the pair constituting 
the internal stress at any point of the bar. Such a force acts 
from, the joint if the stress in the bar is a tension ; toward it if 
the stress is a compression. (Art. 69.) 



ROOF TRUSS WITH VERTICAL LOADS. 



63 



Since these four forces form a system in equilibrium, their 
force polygon must close. This condition will enable us to fully 
determine the magnitudes of the forces, provided all but two are 
known, since the polygon can then be constructed as in Art. 18. 




^j-^o 



We cannot, however, begin with the joint just considered, 
since at first the four forces are all unknown in magnitude. 

If, however, we start with the joint gabh, the polygon of 
forces can be at once drawn. For, reasoning as above, it is 
seen that the portion of the truss immediately surrounding 
this joint is in equilibrium under the action of four forces: 
the reaction in the line ga, the load in the line ab, and the 
internal forces in the lines bh, kg. Of these forces, two (the 
reaction and the load) are completely known ; and it is neces- 
sary only to draw a polygon of which two sides represent the 
known forces and the other two sides are made parallel to the 
members bh, kg. Such a polygon is shown in Fig. 30 ; and 
BH and HG represent in magnitude and direction the forces 
whose lines of action are bh, kg. 



64 GRAPHIC STATICS. 

Evidently, BH and HG represent also the magnitudes (Art. 
63) of the internal stresses in the two members bh and hg. 
The nature of these stresses may be found as follows : Since 
the four forces represented in the polygon GABHG are in 
equilibrium, and since GA, AB are the directions of two of 
them, the directions of the other two must be BH, HG. 
Hence, BH acts toward the joint and HG from it. This 
shows that the stress in bh is a compression, while that in Jig is 
a tension. 

Passing now to the joint bcihb, it is seen that of the four 
forces whose lines of action meet there, two are fully known, 
namely, the load BC acting vertically downward and the inter- 
nal force in hb acting toward the joint (since the stress is com- 
pressive), while the remaining two (viz., the internal forces in 
ci and iJt) are unknown in magnitude and direction. Since, 
however, the unknown forces are but two in number, the force 
polygon can be completely drawn, and is represented by the 
quadrilateral HBCIH. The directions of the 'forces are found 
as in the preceding case, and it is seen that the bars ci and ih 
both sustain compressive stresses. 

The process may be continued by passing to the remaining 
joints in succession, in such order that at each there remain to 
be determined not more than two forces. The complete con- 
struction is shown in Fig. 30. 

It is evident that the loads AB and EF might have been- 
omitted without changing the stresses in any of the truss 
members. For their omission would leave as the complete 
force polygon for external forces BCDEGB, and the two 
reactions would be GB and EG ; but the force diagram would 
be otherwise unchanged. 

The great convenience of the notation adopted is now seen.* 



*This is known as Bow's notation. The notation adopted in Part I involves the same 
idea, but it is not usually employed in works on Graphic Statics, though possessing very 
evident advantages. It was suggested to the writer by its use in certain of Professor 
Eddy's works. 



ROOF TRUSS WITH VERTICAL LOADS. 65 

The line representing the stress in any member is designated 
in the force diagram by letters similar to those which designate 
that member in the truss diagram. The latter is evidently a 
space diagrmn (Art. 11). The force diagram is often called a 
stress diagram, since it shows the values of the internal stresses 
in the truss members. 

89. Reciprocal Figures. — There are always two ways of com- 
pleting the force polygon when two of the forces are known 
only in lines of action. (See Art. 18.) Either way will give 
correct results, but unless a certain way be chosen, it will 
become necessary to repeat certain lines in the stress diagram. 
Thus, if, in Fig. 30, instead of GABHG we draw GABH'Gy 
the lines GH', H'B are not in convenient positions for use in 
the other polygons of which they ought to form sides. The 
lettering of the diagrams will also be complicated. As an aid 
in drawing the lines in the most advantageous positions, it is 
convenient to remember the fundamental property of figures- 
related in such a way as the force and space diagrams shown in 
Fig. 30. 

Such figures are said to be reciprocal with regard to each 
other. The fundamental property of reciprocal figures is that 
for every set of lines intersecting in a point in either figure,, 
there is in the other a set of lines respectively parallel to them 
and forming a closed polygon. 

It is also an aid to remember that the order of the sides in 
any closed polygon in the stress diagram is the same as the 
order of the corresponding lines in the truss diagram, if taken 
consecutively around the joint. This usually enables us at 
once to draw the sides of each force polygon in the proper 
order. 

90. Order of External Forces in Force Polygon. — It will be 
observed that in the case above considered, in constructing the 
force polygon for the loads and reactions, these forces have 
been taken consecutively in the order in which their points of 



66 



GRAPHIC STATICS. 



application occur in the perimeter of the truss. This is a 
necessary precaution in order that the stress diagram and truss 
diagram may be reciprocal figures, so that no line in the former 
need be duplicated. 

This requirement should be especially noticed in such a case 
as that shown in Fig. 31, in which loads are applied at lower as 




well as at upper joints. If the reactions are found by the 
method of Art. 8y, without modification, the force polygon will 
not show the external forces in the proper order, since the 
known forces are not applied at consecutive joints of the truss. 
A new polygon should therefore be drawn after the reactions 
have been determined. 

If desirable (as in some cases it may be) to make use of a 
funicular polygon in which the external forces are taken con- 
secutively, this may be drawn after the reactions are deter- 
mined and the new force polygon is drawn. 

If a load be applied at some joint interior to the truss, as at 
J/ (Fig. 31), then in constructing the stress diagram it should 



STRESSES DUE TO WIND PRESSURE. 6/ 

be assumed to act at N, where its line of action intersects the 
exterior member of the truss, and the fictitious member MN 
inserted. The stresses in the actual truss members will be 
unaffected by this assumption, and such a device is necessary in 
order that the stress diagram may be the true reciprocal of the 
truss diagram. 

§ 3. Stresses Due to Wind Pressure. 

91. Direction of Reactions Due to Wind Pressure. — Since 
the effective pressure of the wind has the direction normal to 
the surface of the roof (Art. 85), it has a horizontal component 
which must be resisted by the reactions at the supports. 
These cannot, therefore, act vertically, as in the case when 
the loading is vertical. Their actual directions will depend 
upon the manner in which the ends of the truss are sup- 
ported. 

If the ends of the truss are immovable, the directions of the 
reactions cannot be determined, since any one of an infinite 
number of pairs of forces acting at the ends would produce 
equilibrium. (The same would be true of the reactions due to 
vertical loads.) In such a case the usual assumption is one of 
the following : (i) the reactions are assumed parallel to the 
loads ; (2) the resolved parts of the reactions in the horizontal 
direction are assumed equal. 

In the case of trusses of large span it is not unusual to 
support one end of the truss upon rollers so that it is free to 
move horizontally, the other end being hinged, or otherwise 
arranged to prevent both horizontal and vertical motion. This 
allows for expansion and contraction with change of tempera- 
ture, as well as for movements due to the small distortions of 
the truss under loads. With this arrangement the reaction at 
the end supported on rollers must be vertical ; and since the 
point of application of the other' reaction is known, both can 
be fully determined by the method described in Art. 39. 



68 GRAPHIC STATICS. 

92. Determination of Reactions. — The methods of finding 
reactions will now be explained for the three cases mentioned in 
the preceding article : (i) Assuming both reactions parallel to the 
wind pressure ; (2) assuming the horizontal resolved parts of the 
two reactions equal ; and (3) assuming one reaction vertical, 

(i) The first case needs no explanation, since it is identical 
with that described in Art. ^y, except that the loads and 
reactions have a direction normal to one surface of the roof,, 
instead of being vertical. It is to be noticed that this assump- 
tion cannot be made if the roof surface is curved, since the 
lines of action of the forces will not be parallel. But since 
the direction of the resultant of the loads will be known from 
the force polygon, both reactions may be assumed to act par- 
allel to this resultant, and the construction made as before. 

(2) In the second case, let each reaction be replaced by two 
forces acting at the support, one horizontal and the other 
vertical. The two horizontal forces are known as soon as the 
force polygon for the loads is drawn, and the two vertical 
forces may be found as in the preceding case, since their lines 
of action are known. 

In Fig. 32, let ab, bcy cd be the lines of action of the wind 
forces. Let the right reaction be considered as made up of a 
horizontal component acting in de and a vertical component 
acting in ef; and let the left reaction be replaced by a vertical 
component acting in fg and a horizontal component acting in 
ga. Draw the force polygon (or " load-line ") ABCD. By the 
assumption already made GA and DE are to be equal, and 
their sum is to equal the horizontal resolved part of AD. 
Through the middle point of AD draw a vertical line ; its 
intersections with horizontal lines through A and D determine 
the points G and E, so that the two forces GA and DE become 
known. The only remaining unknown forces are the vertical 
forces EF 2^\^ FG. Choose a pole O, draw rays' to the points 
G, A, B, C, D, E, and then the corresponding strings. Through 
the points determined by the intersection of og withy^, and oe 



STRESSES DUE TO WIND PRESSURE. 



69 



with ef, draw the string of. The corresponding ray drawn 
from O intersects EG in the point F, thus determining EF and 
FG. The reactions are now wholly known ; that at the left 
support being FA, and that at the right support DF. 



'^r 




7^T 



(3) For the third case the construction is shown in Fig. 33 
(A) and {E), for the two opposite directions of the wind. The 
method is identical with that employed in Art. 39. Only one 
point of the line of action of the left reaction is known, hence 
this is taken as the point of intersection of the corresponding 
strings of the funicular polygon. One of these strings can be 
drawn at once, since the corresponding ray is known ; and the 
other is known after the remaining strings have been drawn, 
since it must close the polygon. The construction should be 
carefully followed through by the student. 

The force polygons for the two directions of the wind are 
distinguished by the use of O and O^ to designate the two poles. 



70 



GRAPHIC STATICS. 



In Fig. 33 (A), ABCDEFGHIA is the force polygon for the 
case when the wind is from the right. Notice that the points 
A, B, C, D, coincide. This means that the loads AB, BC, CD, 
are each zero. 




Fig. 33 P 




^o' 



EFGH 



DABC 




r=>o 



In diagram {B), ABCDEFGHI A is the force polygon io\ 
the case when the wind is from the left. The points E, F, G, H, 
coincide, because the loads EF, EG, GH, are each zero. 



93. Stress Diagrams for Wind Pressure. — When the loads 
and reactions due to wind pressure are known, the internal 
stresses can be found by drawing a stress diagram, just as in 
the case of vertical loads. The construction involves no new 
principle, and will be readily made by the student. In Figs. 
33 {A) and 33 (B) are shown the diagrams for the two directions 
of the wind. 



MAXIMUM STRESSES. 



71 



The stresses in all members of the truss must be determined 
for each direction of the wind. If the truss is symmetrical 
with respect to a vertical line, as is usually the case, it may be 
that the same stress diagram will apply for both directions of 
wind. This will be so if the reactions are assumed to act as 
in cases (i) and (2) of the preceding article. In the case repre- 
sented in Fig. 33, however, the hinging of one end of the truss 
destroys the symmetry of the two stress diagrams, and both 
must be drawn in full. 

§ 4. Maximum Stresses. 

94. General Principles. — For the purpose of designing any 
truss member, it is necessary to know the greatest stresses to 
which it will be subjected under any possible combination of 
loads. 

Stresses are combined in accordance with the principle stated 
in Art. yS, that the resultant stress in a truss member due to 
the combined action of any loads is equal to the algebraic sum 
of the stresses due to their separate action. 

The method will be illustrated by the solution of an example 
with numerical data, 

95. Problem — Numerical Data. — Let it be required to design 
a wrought iron truss of 40 ft. span, of the form shown in Fig. 
34 (PL I). Let 12 ft. be the distance apart of trusses, and let 
the loads be as follows : 

Weight of truss, to be assumed in accordance with the 
formula of Art. 82 : />F=|- «/ (i + Jq- /). This gives PF= 1800 lbs. 
Assuming this to be divided equally among the upper panels, 
and that the load for each panel is borne equally by the two 
adjacent joints, the load at each of the joints be, cd, de is 450 
lbs. The loads at the end joints may be neglected, being borne 
directly at the supports. 

Weight of roof. — This depends upon the materials used and 
the method of construction, but will be taken as 6 lbs. per sq. ft. 



72 



GRAPHIC STATICS. 



•of roof area, giving 900 lbs. as the load at each joint. This, 
also, is a permanent load. Total permanent load per joint, 
1350 lbs. 

Weight of snow. — Taking this as 1 5 lbs. per horizontal 
square foot, we find 1800 lbs. as the load at each joint. 

Wind pressure. — This is computed from the formula 

2 sin a . / \ i. o^ \ 

A= ^^ • . A- (Art. 85. 

I + sm" a 

For this case we put sina = |-| = -|; /„ = 40 lbs. per sq. ft.; 
whence pa. = ZS lbs. per sq. ft. (about). This gives upon each 
panel of the roof 5250 lbs. Then with the wind from either 
.side, the wind loads on that side would be 2625 lbs., 5250 lbs., 
2625 lbs. respectively. 

96. Stress Diagrams. — We are now ready to construct the 
stress diagrams. 

The truss is shown (PL I) in Fig. 34 {A). Fig. 34 {B) is the 
stress diagram for permanent loads. No diagram for snow 
loads is needed, since it would be exactly similar to that for 
permanent loads. The snow load at any joint being four-thirds 
as great as the permanent load, the stress in any member due 
to snow is four-thirds that due to permanent loads. 

Fig. 34 {C) shows the stress diagram for the case of wind 
blowing from the left. The reactions are assumed to act in 
lines parallel to the loads — that is, normal to the roof. With 
this assumption, no separate diagram is needed for the case of 
wind from the right, since such a diagram would be exactly 
symmetrical to Fig. 34 {C). For example, the stress in the 
member gh due to the wind blowing from the right is given by 
the line GM in Fig. 34 {C). 

97. Combination of Stresses. — After the stress diagrams are 
•completed for the various kinds of loads, the- stresses should 
be scaled from the diagrams and entered with proper sign in a 
liable, as follows : 



MAXIMUM STRESSES. 



73 



Member. 


Permanent 
Load. 


Snow. 


Wind R. 


Wind L. 


Max. 


bh 


-4560 


-6080 


— 6270 


-7925 


- 18570 


ci 


-3760 


— 5010 


— 6270 


-7925 


— 16700 


hi 


— 1080 


- 1440 





-5250 


- 7770 


ik 


+ 1785 


+ 2380 


+ 55° 


+ 6650 


+ 10820 


gh 


+ 3735 


+ 4980 


+ 2600 


+ 8800 


+ 17520 


gk 


+ 2190 


+ 2920 


+ 2300 


+ 2300 


+ 7410 


gm 


+ 3735 


+ 4980 


+ S800 


4- 2600 


+17520 


Ik 


+ 1785 


+ 2380 


+ 6650 


+ 550 


+ 10820 


ml 


— 1080 


-1440 


-5250 





- 7770 


dl 


-3760 


— 5010 


-7925 


— 6270 


— 16700 


em 


-4560 


-6080 


-7925 


— 6270 


- 18570 



By combining the results, the maximum stress in each mem- 
ber for any possible condition of loading can be determined. 
The possible combinations of loading are the following : Perma- 
nent load alone ; permanent and snow loads ; permanent load, 
and wind from either direction ; permanent and snow loads, and 
wind from either direction. The student will readily under- 
stand the method of combining the separate results. 

The problem here solved relates to a very simple form of 
truss. With some forms there may occur a reversal of stress 
in certain members, under different conditions of loading. 

It is to be noticed that in the table the word maximum is 
used in its numerical sense, and has no reference to the algebraic 
sign of the stress. 



98. Examples. — In Fig. 35 {A) to {F), are shown several 
forms of truss for which the student may draw stress diagrams, 
assuming loads in accordance with the data given in Arts. 82 
to 85. In determining reactions due to wind pressure, the 



74 



GRAPHIC STATICS. 



three assumptions mentioned in Art. 92 should all be used in 
different cases, that the student may become familiar with the 
principle of each. 




Fif?. 35 



§ 5. Cases Apparently Indeterminate. 

99. Failure of Usual Method. — In attempting to construct 
the stress diagram by drawing the force polygon for each joint 
in succession, as in the cases thus far treated, a difficulty is 
met in certain forms of truss. It may happen that after pro- 
ceeding to a certain point it is impossible to select a joint for 
which the force polygon can be completely drawn, the number 
of unknown forces for every joint being greater than two. 

Thus, in the truss shown in Fig. 36, if the stress diagram is 
started in the usual way, beginning at the left support, the force 
polygons for three joints may be constructed without difficulty, 
thus determining the stresses in bl, Ik, hn, cm, mn, nk. But 
the force polygon for cdqpnmc cannot be constructed, since 
three forces are unknown, — namely, those in dq, qp, pn. And 
at the joint knpsk, the stresses in np, ps, and sk are unknown. 
The problem, therefore, seems at this point to become indeter- 



CASES APPARENTLY INDETERMINATE. 



75 



minate, since either of the two polygons can be completed in 
any number of ways, so far as the known forces determine. It 
can be shown, however, that this ambiguity is only apparent. 
This may be proved as follows : 

Consider the portion of the truss to the left of the broken 
line M'N'. It is in equilibrium under the action of eight 
forces ; five of these (four loads and a reaction) are known ; 
the remaining three are the forces in er, rs, and sk. Now, the 
problem of determining these three unknown forces is the same 
as that treated in Art. 40. It was there found to be a deter- 
minate problem, unless the lines of action of the three unknown 
forces intersect in a point or are parallel. 

That the problem is determinate may be seen also from the 
principle of moments (Art. 51). The eight forces mentioned 
being in equilibrium, the sum of their moments is zero for any 
origin in their plane. Let the origin be taken at the point of 
intersection of the lines of action of two of the unknown forces, 
as er, rs. Then from the principle of moments we have (since 
the moments of the two forces named are zero) : Algebraic sum 
of moments of loads and reaction to left of section + moment 
of SK=o. The only unknown quantity in this equation is the 
magnitude of SK, which may, therefore, be determined. The 
other unknown forces may be found in a similar manner, 
the origin of moments being in each case chosen so as to 
eliminate two of the three unknown forces. 

The whole problem of drawing the stress diagram is now 
seen to be determinate. For, as soon as the stress in sk is 
known, the force polygon for the joint knpsk contains but two 
unknown sides, and can be drawn at once. No further diffi- 
culty will be met. 

100. Solution of Case of Failure — First Method. — The 

reasoning of the preceding article suggests two methods of 
treating the so-called ambiguous case. These will now be 
described. 



^6 



GRAPHIC STATICS. 



The first method is to apply the construction of Art. 40, as 
follows : Referring to Fig. 36, consider the equilibrium of the 
portion of the truss to the left of the line M'N'. The system 
of forces consists of those whose lines of action are ka, ab, be, 
cd, de, er, rs, sk. Let them be taken in the order named, and 
draw the force polygon for the known forces. (The reaction 
ka is supposed to be already determined.) The known part of 
the force polygon is KABCDE', the unknown part is to be 
marked ERSK. Choose a pole O and draw rays to K, A, B, 
C, D, and E ; then draw the corresponding strings of the funic- 
ular polygon. Remembering the method of Art. 40, we draw 
first oe, making it pass through the intersection of er and rs 




(the reason for this being that it makes the string os also pass 
through that point) ; then draw in succession od, oc, ob, oa, ok. 
The string ok intersects sk in a point through which os must be 
drawn. Hence os must join that point with the starting point 
of the polygon. This completes the funicular polygon, except 
the string or, which must pass through the intersection of er 



CASES APPARENTLY INDETERMINATE. 



77 



and rs in some direction not yet known. This string is not 
necessary to the solution of the problem. 

Since os is now known, OS may be drawn parallel to it from 
O ; and by drawing a line from K parallel to the known direc- 
tion of KS, the point 5 is determined, and KS becomes 
known. 

To find ER and RS it is only necessary to draw from 5 a 
line parallel to rs, and from E a line parallel to er ; their inter- 
section gives the point R, and the force polygon for the system 
of forces considered is complete. The stresses in er, rs, and sk 
being now known, the stress diagram may be completely drawn 
by the usual method. 

It is interesting to notice that the method just described 
determines the lines ER, RS, SK, in their proper position in 
the complete stress diagram. The determination of ER and 
i?5 by this method is not necessary, since the usual method of 
drawing the stress-diagram can be carried out as soon as SK 
is known. 

The funicular polygon employed in the above construction, 
so far as it belongs to the external forces, may coincide with 
the corresponding part of the funicular polygon used in deter- 
mining the reactions. If this is desired, two points must 
be observed: (i) the string oe must be the first drawn, and 
must pass through the intersection of er and rs, and (2) the 
pole must be so chosen that ok will not be nearly parallel 
to ks. 

It should also be noticed that the construction of the funic- 
ular polygon might begin with the string ok, which should then 
be made to pass through the intersection of rs and sk. The 
student will be able to carry out this construction without 
difficulty. 

loi. Solution of Case of Failure — Second Method. — It will 
now be shown how the apparently ambiguous case can be 



78 



GRAPHIC STATICS. 



treated graphically by the principle of moments. Referring 
again to Fig. 36, consider the portion of the truss to the left 
of section M'N'. It is acted upon by eight forces, of which 
five (the loads and the reaction) are known, and three (whose 
lines of action are er, rs, sk) are unknown. The algebraic sum 
of the moments of all these forces about any origin must be 
zero. Let the origin be taken at the point of intersection of 
er and rs, so that the moments of the forces acting in these 
lines are both zero ; then the sum of the moments of the five 
known forces, plus the moment of the force acting in the line 
sk, must equal zero. Now the sum of the moments of the 
five known forces may be found by the method of Art. 56. 
Through the origin of moments draw a line parallel to the 
resultant of the forces named (that is, a vertical line), and let 
i equal the length intercepted on it by the strings oe, ok. Then 
the required moment is —iH, where //is the pole distance. 
(The minus sign is given in accordance with the convention 
that left-handed rotation shall be positive.) Let /* = unknown 
force in line sk, and h its moment-arm. For the purpose of 
computing the moment, assume the stress in sk to be a tension ; 
then the force P acts toward the right and its moment is posi- 
tive, the value being + jP/^. 

H ence, Ph — Hi — o ; 

or, P = -H. 

Ji 

From this equation P may be computed. The computation 
may be made graphically as follows : Draw (Fig. 36) a triangle 
WUV, making WU=H (force units) and WV=h (linear 
units). Lay off WY=i (linear units) and draw YX parallel 
to VU. Then WX (force units) represents P. This is readily 

seen, since from the two similar triangles we have the propor- 

P i . 
tion — - = -, which agrees with the equation above deduced. 
H h 

The computation is simplified if the pole distance H is taken 
equal to as many force units as h is linear units ; or if //" 



CASES APPARENTLY INDETERMINATE. 



79 



is some simple multiple of h. For, suppose H=nh\ then 

p = l!l/L=^ni. If n=i, P = i. 
h 

The stress in sk is found to be a tension, since P is positive. 
Whatever the nature of the stress, it may be assumed a ten- 
sion in writing the equation, and the sign of the value found 
will show whether the assumption coincides with the fact. 

I02. Other Methods for Case of Failure. — In certain cases 
the method of treating the " ambiguous case " may profitably 
be varied. 

(i) The construction of Art. lOO may be modified as fol- 
lows : Determine the resultant of the known external forces 
acting on the portion of the truss to one side of the section 
M'N'. This resultant is in equilibrium with the three unknown 
forces acting in the members er, rs, sk. Hence these forces can 
be determined by the special method explained in Art. 42. 

The resultant of the five known forces is represented in 
magnitude and direction by KE in the force polygon ; and its 
line of action passes through the intersection of the strings oe, 
ok in the funicular polygon. Since this point of intersection is 
likely to be inaccessible, the construction of Art. 42 cannot be 
conveniently applied. It may be modified by using instead of 
KE the two forces KA (the reaction in the line kd) and AE 
(the resultant of the four loads, its line of action being deter- 
mined by the intersection of the strings oa and oe). First 
determine forces in the three lines er, rs, sk, which shall be in 
equilibrium with KA ; then make a similar construction for 
AE, and combine the results. 

(2) It has been proposed to employ the following reasoning : 
Remove the members pq and qr, and insert another represented 
by the broken line in Fig. 36. Evidently this does not change 
the stress in the member sk, since the forces acting on the 
truss to the left of the section M^N' are unchanged. But with 
this change the difficulty encountered in constructing the stress 
diagram by the usual method is avoided. For when the joint 



8o GRAPHIC STATICS. 

nmcdqpit is reached, the forces acting there will be all known 
except two. Let the stress diagram be drawn in the usual way 
until the stress in sk is known. Then restore the original 
bracing and repeat the construction, using the value just deter- 
mined for SK. 

This method is convenient whenever it is applicable. Cases 
may, however, arise, in which it will fail. For instance, if 
a load is applied at the joint pqrsp, the members pq and qr 
cannot both be omitted, and the method cannot be applied. In 
such cases, one of the methods explained in the preceding 
articles may be applied. 

Other methods might be mentioned, but the foregoing dis- 
cussion of the case will probably be found sufficient. 

103. Failing Case in Other Forms of Truss. — The usual 
method of constructing the stress diagram may fail in other 
forms of truss, though the one above described is the most 
common. In such a case, the problem of finding the stresses 
may be really indeterminate, or only apparently so. Whenever 
it is possible to divide the truss into two parts by cutting three 
members which are not parallel and do not intersect in one 
point, the stresses in the three members cut are determinate as 
soon as all external forces are known, and can be found by 
methods already given. If more than three members are cut, 
the problem of finding the stresses in them is indeterminate, 
unless all but three of these stresses are known. By remem- 
bering these principles, the determinateness of any given prob- 
lem may readily be tested. (See Art. 70.) 

§ 6. Three-Hiiiged Arch. 

104. Arched Truss Defined. — If a truss is so supported that 
when sustaining vertical loads the reactions at the supports 
have horizontal components directed toward the center of the 



THREE-HINGED ARCH. 8 1 

span, the truss becomes an arcJi. The only kind of arch we 
shall here consider is that consisting of two partial trusses 
hinged together at the crown, and each hinged at the point of 
support. 

Such a truss is shown in Fig. "X,^, in which the two partial 
trusses are hinged to the abutments at P and Q, and connected 
by a hinge at the point R. Since a hinge at the support allows 
the reaction of the supporting body upon the truss to take any 
direction in the plane of the truss, the directions of the reactions 
at P and Q are unknown, as is also that of the force exerted by 
either partial truss upon the other at the point R. The problem 
of determining the reactions may, therefore, at first sight seem 
indeterminate. It will be shown in the next article that it is 
in reality determinate, and that the only principles needed in 
the solution are such as have been already often applied in the 
preceding chapters. The three-hinged arch is indeed a "simple" 
structure (Art. 75), since the theory of elasticity is not needed 
in the determination of the reactions. 

105. Reactions Due to a Single Load. — The method of find- 
ing the reactions is most clearly understood by considering the 




effect of a single load on either partial truss. Let a load be 
applied at 6" (Fig. 38) and let all other loads acting on either 
portion of the truss (including the weight of the structure) be 



;82 GRAPHIC STATICS. 

neglected. Call the two partial trusses X and Y, and consider 
the part Y. The only forces acting upon it are the reaction 
exerted by the abutment at Q and the force exerted at R by 
the truss X. These two forces, being in equilibrium, must 
have the same line of action, which is, therefore, the line QR. 
Consider now the body X. The external forces acting upon it 
are the load at S, the reaction of the abutment at Z', and the 
force exerted by Y at the point R. But this last force is equal 
and opposite to the force exerted by X upon Y, and its line of 
action is therefore QR. The three forces acting upon the 
body X being in equilibrium, their lines of action must meet in 
a point ; which point is found by prolonging QR to meet the 
line of action of the applied load. Let T be this point, then 
RT IS the line of action of the reaction at P. The reactions 
■can now be determined by drawing the triangle of forces. This 
triangle is ABC in the figure, AB representing the load at vS, BC 
the reaction in the line QR (also marked be), and CA the reac- 
tion in the line PT (marked also ca). Evidently ABC may be 
regarded as the polygon of external forces, either for the partial 
truss X, or for the whole structure composed of X and Y; and 
BC represents either the force exerted by Y upon X at R, or 
the force exerted upon Y by the supporting body at Q. 

If, now, the structure be loaded at other points, the reactions 
due to each load may be found separately ; the resultant of 
all such separate reactions at either support will be 'the true 
reaction at that support when all the loads act together. A 
convenient method of applying these principles will be given in 
the next article. 

1 06. Reactions and Stresses Due to Any Vertical Loads. — In 

Fig. 39 is represented a truss consisting of two parts supported 
by hinges at P' and Q' and hinged together at R' . Consider 
all vertical loads to be applied at the upper joints, their lines of 
action being marked in the usual way. We shall first explain 
the construction for finding the reactions at the supports ; after 



THREE-HINGED ARCH. 



83 



these are determined, the drawing of the stress diagram will 
present no difficulty. 

Since we shall sometimes deal with one of the partial trusses, 
and sometimes with the two considered as a single body, it will 
be well at the outset to specify the external forces acting on 
each of these bodies. 




Fi"@-30 



(i) For the partial truss at the left we have five applied loads, 
the reaction at R' (exerted by the other truss), and the reaction 
at P' . The force polygon for these forces will be marked as 
follows : ABCDEFLA. (The meaning of the letters will be 
understood before the force polygon is actually drawn, by 
reference to the corresponding letters in the space diagram.) 



84 GRAPHIC STATICS. 

(2) For the right-hand partial truss the external forces are 
five loads and two reactions, and the force polygon will be 
marked thus : FGHIJKLF. (Notice that FL and LF are equal 
and opposite forces, being the "action and reaction" between 
the two trusses at R' ) 

(3) For the combined structure the external forces are the 
ten loads and the reactions at P' and Q . (The action and 
reaction at R' become now internal forces.) The force polygon 
will be marked thus : ABCDEFGHIJKLA. 

Begin the construction by drawing the force polygon for the 
ten loads on the whole structure, lettering it as just indicated. 
Choose a pole O, draw the rays, and then the funicular polygon 
as far as possible. Now consider the right partial truss as 
unloaded. The resultant of the remaining five loads is repre- 
sented in magnitude and direction by AF\ its line of action 
must pass through the intersection of oa and of, and is therefore 
the line marked af. Now, reasoning as in the preceding article, 
we see that the reaction at Q' must act in the line Q R' . Let 
Q R^ intersect af in T' , then P' V is the line of action of 
the reaction at P'. Hence the complete force polygon for the 
whole truss, when the right half is unloaded, may be found by 
drawing from i^a line parallel to Q R' , and from A a line parallel 
to P' T' , prolonging them till they intersect at L' . The reaction 
at P' is L'A, and that at Q' is FL'. (The line of action of the 
latter is marked//'.) 

Next, consider the left half to be without loads, the other 
half being loaded. The resultant of the five loads now acting is 
FK, its line of action fk being drawn through the point of 
intersection of of and ok. The reactions at P' and Q' for the 
present case have lines of action P' R' and Q' T" , found just 
as in the first case of loading. These reactions are therefore 
determined in magnitude and direction by drawing from K a 
line parallel to Q'T" and from F a line parallel to P'R', pro- 
longing them till they intersect at L" . The complete force 
polygon for this case of loading is therefore FGHIfKL"F. 



THREE-HINGED ARCH. 



85 



Consider now that both parts of the truss are loaded. The 
reaction at P' is the resultant of the two partial reactions V A, 
L"F, and the reaction at Q' is the resultant of the two partial 
reactions FL' , KL". From L" and L' draw lines parallel 
respectively to FL' and FL" , intersecting in L. Then L" L is 
equal and parallel to FL' , and LL' is equal and parallel to L"F. 
Hence JTZ. and LA represent the resultant reactions at Q' and 
P' respectively. This completes the polygon of external forces 
for the whole truss, as well as that for each partial truss. 

The stress diagram can now be drawn in the usual way, 
beginning at the point P' . The diagram for one partial truss 
is shown in Fig. 39 (B). 

If loads are applied at the lower joints of the trusses, the 
reactions due to these may be found in the same manner as for 
the upper loads. But before beginning the determination of 
the stresses, the polygon must be drawn for the external forces 
taken in order around the truss. (See Art. 90.) 

107. Case of Symmetrical Loading. — If the two half trusses 
are exactly similar and symmetrically loaded, the determination 
of reactions and stresses is much simplified. 

(i) As regards the reactions, symmetry shows that the forces 
exerted by the two trusses upon each other at R' are horizontal. 
Hence, referring to Fig. 39, and considering either half-truss, 
as that to the left, the line of action of the reaction at P' may 
be found by drawing a horizontal line through R' and prolong- 
ing it to intersect af, the line of action of the resultant of all 
loads on the left truss ; the line joining this point of intersection 
with P' is the required line of action of the reaction at the left 
abutment. The two reactions are now determined by drawing 
from Fz. horizontal line and from A a line parallel to the line 
just determined, and prolonging them till they intersect. 

(2) As to the stresses, only one partial truss need be consid- 
ered, since the stress diagrams for the two portions will be 
symmetrical figures. 



86 



GRAPHIC STATICS. 



These principles might have been employed in the case dis- 
cussed in the preceding article ; but the method there used i& 
applicable to cases in which either the trusses or the loads are 
unsymmetrical. 

1 08. Wind Pressure Diagram. — The diagrams for wind 
pressure will present no difficulty. The determination of the 
reactions will, indeed, be simpler than in the preceding case, 
since only one partial truss will be loaded at any one time, and 
the line of action of one reaction is therefore known at the 
outset. The construction is shown in Fig. 40. 



J/k 




Fig. 40. 



In computing wind pressure loads, it will be assumed that 
each joint sustains half the pressure coming upon each of the 
two adjacent panels. It will be sufficiently correct in comput- 
ing the pressure at any joint, to assume the slope of the roof 



THREE-HINGED ARCH. 



8/ 



as that of the tangent to the roof curve at the joint in ques- 
tion ; and the direction of the wind load may be taken as that 
of the normal to the roof curve at the joint. The force poly- 
gon for the wind loads is therefore not a straight line when the 
roof is curved. In Fig. 40, this polygon is FGHIJK. The 
reactions are to be marked KL, LF. (Since there are no loads 
on the other half-truss, the points ABCDEF will coincide.) 
Choosing a pole O, draw the funicular polygon as shown, and 
determine the line of action of the resultant of all the wind 
loads. This line of action/^ is drawn parallel to FK, through 
the point of intersection of the strings of and ok. Prolong //& 
to intersect P^K produced at P' ; then g'P' is the line of 
action of the reaction at Q. The force polygon may now be 
completed by drawing KL parallel to QV and LF parallel to 
P^R\ The reactions KL and LF being thus determined, the 
stress diagram can be drawn without difficulty. 

The stress diagram is drawn for both partial trusses, with the 
result shown in the iigure. If the two trusses are symmetrical, 
the diagram for the other direction of the wind need not be 
drawn ; the stresses for this case being found from the diagram 
already drawn. If, however, the partial trusses are dissimilar, a 
second wind-diagram must be drawn. 

It is not necessary to draw separate space diagrams for verti- 
cal loads and wind-forces. The constructions of Figs. 39 and 
40 have been here kept wholly separate, in order that the 
explanation may be more easily followed. 

109. Check by Method of Sections. — In case of a truss of 
long span, especially when the members have many different 
directions and are short compared with the whole length of the 
span, the small errors made in drawing the stress diagram are 
likely to accumulate so much as to vitiate the results. Thus, 
in Fig. 39, if, in drawing the stress diagram, we begin at P' and 
pass from joint to joint, there is no check upon the correctness 
of the work until the point R^ is reached. At that point we 



88 GRAPHIC STATICS. 

have a second determination of the reaction exerted by each 
half-truss upon the other ; and it is quite likely that the two 
values found will not agree. 

By a method like that employed in Art. loo for the "inde- 
terminate " form of truss, we may avoid the necessity of mak- 
ing so long a construction before checking the results. In 
Fig. 39 take a section cutting the three members cq, qr, rl, 
and apply the principles of equilibrium to the body at the 
left of the section. The forces acting on this body are LA, 
AB, BC, CQ, QR, RL. Choose a pole O' and draw the funic- 
ular polygon for this system, making the string o'c pass through 
the point of intersection of eg and qr. Draw successively the 
strings o'c, o'b, o^a, o'l, prolonging the last to intersect Ir. 
Through the point thus determined, draw o'r, which must also 
pass through the point of intersection of cq and qr. As soon as 
v'r is known, the corresponding ray O'R can be drawn in the 
force diagram, and then by drawing from L a line parallel to 
Ir, the point R is determined. We may now close the force 
polygon, since the directions of the two remaining forces [cq 
and qj) are known. 

The stresses in the three members cut being now known, we 
have a check on the correctness of the construction of the 
stress diagram, as soon as these members are reached in the 
process. 

This method will be found of great use, not only for this 
form of truss, but for any truss containing many members. 

§ 7. Coiinterbracing. 

no. Reversal of Stress. — If the loads supported by a truss 
are fixed in position and unchanging in amount, the stress in 
any member remains constant in magnitude and kind. But in 
most cases such are not the conditions, and it may happen that 
under different combinations of loading, the stress in a certain 
member is sometimes tension and sometimes compression. It 



COUNTERBRACING. 



89 



is often thought desirable to prevent such changes of stress, 
the design of the members and their connections being thereby 
simplified. To accomplish this is the object of coiinterbracing. 

III. Counterbracing. — Consider a truss such as the one 
shown in Fig. 41, subjected to vertical loads and to wind pres- 
sure from either side. A diagonal member such as xy may 
sustain tension under vertical loads alone, or with the wind 
from the left ; while with the wind from the right it may sustain 
compression. 

Now suppose xy removed, and a member represented by the 
broken line xy' introduced. It may easily be shown that any 




system of loading which would cause compression in xy will 
cause tension in this new member ; and vice versa. For, divide 
the truss by a section MN cutting xy and two chord mem- 
bers as shown, and let L be the point of intersection of the 
two chord members produced. The kind of stress in xy or 
the other member (whichever is assumed to be present) may 
be determined by considering the system of forces acting on 
■one portion of the truss, as that to the left of the section MN. 
Let the principle of moments be applied to this system, the 
origin being taken at L. If the external forces acting on the 
portion of the truss considered be such as to tend to cause 
right-handed rotation about L, the stress in xy must be com- 
pression in order to resist this tendency ; while, if xy be replaced 
by xy , a tension must exist in that member to resist right- 
handed rotation about L. Similarly, a tendency of the external 
forces to produce left-handed rotation about L would be resisted 
l)y a tension in xy ; or by a compression in the member xy\ 
If the two members act at the same time, the stresses in 



90 



GRAPHIC STATICS. 



them will be indeterminate. These stresses may, however, be 
made determinate by the following device : 

Let the member xy be so constructed that it cannot sustain 
compression. Then, whenever the external forces are such as 
to tend to throw compression upon it, it ceases to act as a 
truss-member, and the member xf receives a tension which is 
determinate. 

If the member xy' be constructed in the same way, any 
tendency to throw compression upon it causes it immediately 
to cease to act, and puts upon the member xy a tension instead. 

A member such as xy\ constructed in the manner mentioned, 
is called a coimterbrace. 

112. Determination of Stresses with Counterbracing. — The 

use of counterbracing adds somewhat to the labor of deter- 
mining the maximum stresses, since the members actually under 
stress are not always the same. The method of treating such 
cases will be illustrated in the next article by the solution of an 
example ; but first the main steps in the process may be outlined 
as follows : 

{a) Construct separate stress diagrams for vertical loads and 
for wind in each direction, assuming the diagonals in all panels 
to slope the same way. 

ib) Determine by comparison of these diagrams in which of 
the diagonal members the resultant stress is ever liable to be a 
compression. Draw in counters to all such members. 

{c) With these counterbraces substituted for the original 
members, either draw new stress diagrams, or make the neces- 
sary additions to those already drawn. If the latter method is 
adopted, the added lines should be inked in a different color 
from that employed originally. (In some cases, this construc- 
tion may be unnecessary on account of symmetry, as will be 
illustrated in the next article.) 

{d) Combine the separate stresses for maxima in the usual 
way. 



COUNTERBRACING. 



91 



113. Example. — In Fig. 42 (PI. II) are given stress diagrams 
for a " bow-string " roof-truss in which counterbracing is em- 
ployed. At (A) is shown the truss or space diagram. The 
span is 48 ft. ; rise of top chord, 16 ft. ; rise of bottom chord, 
8 ft. The chords are arcs of parabolas. The whole truss is 
divided into six panels by equidistant vertical members. The 
distance apart of trusses is taken as 12 ft. 

Assume the weight of the truss at 2400 lbs., and that of the 
roof at 3600 lbs. ; this makes the total permanent load 1000 lbs. 
per panel. Take 800 lbs. as the load at each upper joint, and 
200 lbs. as the load at each lower joint. 

The snow load, computed at 15 lbs. per horizontal square 
foot, is about 1440 lbs. per panel. 

Wind pressure is to be computed from the formula of Art. 85. 

We now proceed to apply the method outlined in the preced- 
ing article. 

(a) Assuming one set of diagonals present, we construct the 
stress diagrams for the various kinds of loads. 

Diagram for permanent loads. — This is shown at {B) Fig. 
42, which needs little explanation. It will be noticed that the 
stress in every diagonal member is zero. This will always be 
the case if the chords are parabolic and the vertical loads are 
equal and spaced at equal horizontal distances. 

Diagram for snow loads. — Fig. 42 iC) is the stress diagram 
for snow loads. In this case, also, the stresses in the diagonals 
are all zero. 

Wind diagrajns. — At {D) and {E) are shown the diagrams 
for the two directions of the wind. The only thing needing 
special mention is the method used in laying off the force- 
polygon for the wind loads. We first compute the normal wind 
pressure on each panel by the formula of Art. 85. We thus 
find, when the wind is from the right, the following total pres- 
sures, taking /„ = 40 lbs. per sq. ft. : 

On panel d, 1650 lbs. On panel e, 3870 lbs. On panel/, 5480 lbs. 



92 



GRAPHIC STATICS. 



In Fig. 42 {D) these are laid off successively in their proper 
directions to the assumed scale. Thus CD', D'E', E'G repre- 
sent respectively 1650 lbs. normal to dq, 3870 lbs. normal to es, 
and 5480 lbs. normal to ft. Now each of these loads is to be 
equally divided between the two adjacent joints. Bisect CU at 
D, D'E' at E, and E'G at F; then CD, DE, EF, EG represent 
the loads at the joints cd, de, ef, and fg respectively. The 
"load line" is therefore CDEFG. 

The reactions are assumed to be parallel to the resultant 
load. With this explanation the figures {D) and {E) will be 
readily understood. 

ip) Comparison of results. — The stresses due to permanent 
loads, wind right, and wind left, are shown in tabular form for 
convenience of comparison. 



Member. 


Perm. Load. 


Snow Load. 


Wind R. 


Wind L. 


Max. 


aj 


-6660 


-9680 





4250 


_ 


14070 


- 16340 


bl 


-5350 


- 7780 


- 


4880 


- 


9650 


- I3130 


en 


-4550 


— 6640 


- 


6380 


- 


6200 


— 1 1 190 


dq 


-4550 


— 6640 


- 


9500 


- 


4250 


— 14050* 


es 


-5350 


-7780 


- 


15030 


- 


3450 


— 20380* 


fl 


-6660 


- 9680 


- 


14070 


- 


4300 


— 20730* 


A 


+ 5080 


+ 7400 


+ 


780 


+ 


13500 


+ 12480 


kK 


+ 4680 


+ 6820 


+ 


700 


+ 


12450 


+ 1 1 500 


mh^ 


+ 4470 


+ 6520 


+ 


1950 


+ 


7350 


+ 10990 


Ph 


+ 4470 


+ 6520 


+ 


4100 


+ 


4100 


+ 10990* 


rh.^ 


+ 4680 


+ 6820 


+ 


7680 


+ 


2050 


+ 12360* 


th^ 


+ 5080 


+ 7400 


+ 


13500 


+ 


800 


+ 18580* 


jk 


+ 1 180 


+ 1440 


+ 


150 


+ 


2650 


+ 2620 


Im 


+ 1 180 


+ 1440 


- 


280 


+ 


4100 


+ 2620 


np 


+ I180 


+ 1440 


- 


950 


+ 


3800 


+ 2620* 


qr 


+ 1 180 


+ 1440 


- 


1650 


+ 


2525 


r + 2620* \ 
I - 470* i 


St 


+ 1 180 


+ 1440 


- 


1350 


+ 


1250 


/ + 2620* ) 
I- 170*) 


kl 








+ 


1300 


- 


4600 


+ I 300* 


mn 








+ 


2700 


- 


4100 


+ 2700* 


Pi 








+ 


4850 


- 


3150 


+ 4850* 


rs 








+ 


7080 


— 


1950 


+ 7080* 



COUNTERBRACING. q^ 

It is seen that the diagonals shown are all in tension when 
the wind is from the right, and all in compression when the 
wind is from the left. Therefore counters are needed in all 
panels, and the counters will all come in action whenever the 
wind is from the left. 

(c) Stresses in comiterbraces. — It is evident from symmetry 
that no new diagrams are needed to determine the stresses in 
the counterbraces. In fact, the counterbrace in each panel is 
situated symmetrically to the main diagonal in another panel, 
and is subject to exactly equal stresses. 

id) Covibination for maxima. — In combining the results for 
the greatest stresses in the various members, it will be assumed 
that the greatest snow load and the greatest wind load can 
never act simultaneously. For each member, therefore, the 
stress due to permanent load is to be added to the snow stress 
or the wind stress, whichever is greater. Again, in combining 
the tabulated results, we consider only the columns headed per- 
manent load, snow load, and wind right ; since whenever the 
wind is from the left, the other system of diagonals is in action. 
This gives the results entered in the last column. 

We now notice the following facts : 

(i) The results given for the diagonal members are the true 
maximum stresses. 

(2) The stress found for each diagonal applies also to the 
symmetrically situated counterbrace. 

(3) For any other member, we are to choose between the 
maximum found for that member and the value found for the 
symmetrically situated member. Thus, —20730 is the true 
maximum stress for both «y and /^; etc. (The numbers denot- 
ing true maximum stresses are marked with a (*) in the table.) 

It is seen that the verticals, with one exception, may sustain 
a reversal of stress. Thus, 7'/^ and st must be designed for a 
tension of 2620 lbs., and also for a compression of 170 lbs. ; 
while Im and qr are each liable to 2620 lbs. tension and to 470 
lbs. compression. 



CHAPTER VI. SIMPLE BEAMS. 

§ I. General Principles. 

114. Classification of Beams. — A beam has been defined in 
Art. 79. Beams may be treated in two main classes, the basis 
of classification being that described in Art. 75. These two 
classes will be called simple and noji-siniple beams respectively. 
The present chapter deals only with simple beams, the defini- 
tion of which may be stated as follows : 

A simple beam is one so supported that it may be regarded 
as a rigid body in determining the reactions. 

A simple beam may rest on two supports at the ends ; or it 
may overhang one or both supports. 

A cantilever is any beam projecting beyond its supports. 
Such a beam may be either simple or non-simple. 

A cojttinnoiis beam is one resting on more than two" supports. 
Such a beam is non-simple. 

A beam may be supported in several ways. It is simply 
supported at a point when it rests against the support so that 
the reaction has a fixed direction. It is constrained at a point 
if so held that the tangent to the axis of the beam at that 
point must maintain a fixed direction. If hinged at a support, 
the reaction may have any direction. We shall deal mainly 
with the case of simple support. 

In what follows it will be assumed that the beam rests in a 
horizontal position, since this is the usual case.. 

115. External Shear, Resisting Shear, and Shearing Stress. 

— The external shear at any section of a beam is the algebraic 

94 



GENERAL PRINCIPLES. 



95 



sum of the external vertical forces acting on the portion of the 
"beam to the left of the section. 

The resisting shear at any section is the algebraic sum of the 
internal vertical forces in the section acting on the portion of 
the beam to the left, and exerted by the portion to the right 
of the section. 

The sJiearing stress at any section is the stress which con- 
sists of the internal vertical forces in the section, exerted by 
the two portions of the beam upon each other. It consists of 
the resisting shear and the reaction to it. (See Art. 63.) 

Let AB (Fig. 43) be a beam in equilibrium under the action 
of any external forces. At any point in its length, as C, con- 
ceive a plane to be passed perpen- 
dicular to the axis of the beam, and \ ^ ^ 

consider the portion AC, to the left ^^1^.43 

of the section. The principles of equilibrium apply to the body 
AC, and the external forces acting upon it include, besides 
those forces to the left of C that are external to the whole bar, 
certain forces acting across the section at C that are internal to 
AB, but external to AC. (Art. 61.) These latter forces com- 
prise that constituent of the internal stress between AC and 
CB which is exerted by CB upon AC 

Represent by Fthe algebraic sum of the resolved parts in 
the vertical direction of all forces acting on AB to the left of 
the section at C, upward forces being called positive. V is the 
external shear at the given section as above defined. 

Since the body ^C is in equilibrium, condition (i), Art. 58, 
requires that the algebraic sum of the resolved parts in the 
vertical direction of all forces acting on it must equal zero. 
Hence the forces acting on AC in the section at C must have a 
vertical component equal to —V. This vertical component is 
called the resisting shear in the given section. This resisting 
shear is one of the forces of a stress of which the other is an 
■equal and opposite force exerted by /iC' upon CB. This stress 
is called the shearing stress in the section, and is called positive 



96 GRAPHIC STATICS. 

when it resists a tendency oi AC to move upward, and of CB 
to move downward. 



1 1 6. Bending Moment, Resisting Moment, and Stress Moment. 

— The bending 7noinent at any section of a beam is the algebraic 
sum of the moments of all the external forces acting on the 
portion of the beam to the left of the section ; the origin of 
moments being taken in the section. 

The resisting moment at any section is the algebraic sum of 
the moments of the internal forces in the section acting on the 
portion of the beam to the left, and exerted by the portion to 
the right of the section ; the origin of moments being the same 
as for bending moment. 

The stress moment or mom,ent of internal stress at any section 
consists of the two equal and opposite moments of the forces 
exerted across the section by the two portions of the beam 
upon each other. 

Referring again to Fig. 43, let us analyze further the forces in 
the section at C. Applying to the body AC\.\i^ second condition 
of equilibrium ((2) of Art. 58), and taking an origin at a point 
in the section, we see that the algebraic sum of the moments 
of all the external forces acting on the beam to the left of the 
section plus the sum of the moments of the internal forces act- 
ing on AC in the section must equal zero. The former sum is 
defined as the bending momejit at the given section. Represent 
it by M. The latter sum is defined as the resisting moment at 
the section, and must be equal to —M, by the above principle. 

We have thus far referred only to the internal forces exerted 
by CB upon AC\ but evidently the equal and opposite forces 
exerted hy AC upon CB have a moment numerically equal to 
M. The equal and opposite moments of the equal and opposite 
forces of the stress in the section together constitute the stress 
moment in the section. 

If the external forces applied to the beam are all vertical, the 
value of M will be the same at whatever point of the section 



GENERAL PRINCIPLES. 



9T 



the origin is taken ; since the arm of each force will be the same 
for all origins in the same vertical line. If the loads and reactions 
are not all vertical, the value of M will generally depend upon 
what point in the section is taken as the origin of moments. 

117. Curves of Shear and Bending Moment. — The curve of 
shear for a beam is a curve whose abscissas are parallel to the 
axis of the beam, and whose ordinate at any point represents 
the external shear at the corresponding section of the beam. 

Let AB (Fig. 44) represent a beam loaded in any manner, and 
let A^B^ be taken parallel to AB. At every point of A^B^ 
suppose an ordinate drawn whose length shall represent the 
external shear at the corresponding g 

section of AB. The line ab, join- 
ing the extremities of all these ordi- 
nates, is the shear curve. Positive 
values of the shear may be repre- 
sented by ordinates drawn upward 
from AB\ and negative values by --«.--^ 

ordinates drawn downward. (Instead of drawing AB^ parallel 
to the beam, it may be any other straight line whose extremi- 
ties are in vertical lines through A and B) 

The curve of bending moment for a beam is a curve whose 
abscissas are parallel to the axis of the beam, and whose ordinate 
at any point represents the bending moment at the correspond- 
ing section of the beam. Thus in Fig. 44, A"C"B" may repre- 
sent the bending moment curve for the beam AB. (Evidently 
A"B" might be inclined to the direction of AB, without destroy- 
ing the meaning of the curve.) Positive and negative values of 
the bending moment will be distinguished by drawing the 
ordinates representing the former upward and those represent- 
ing the latter downward from the line of reference A"B". 

118. Moment Curve a Funicular Polygon. — If the loads and 
reactions upon the beam are all vertical, every funicular poly- 
gon for these forces taken consecutively along the beam, is a. 




98 



GRAPHIC STATICS. 



curve of bending moments. Thus, let MN (Fig. 45) represent 
a beam under vertical loads, supported at the ends by vertical 
reactions. Draw a funicular polygon for the loads and reactions 
as shown. 



c\d N 




-:::^o 



Fig. 45 

Now, by definition, the bending moment at any section is 
equal to the moment of the resultant of all external forces act- 
ing on the beam to the left of the section, the origin of 
moments being taken in the section. By Art. 56, this moment 
can be found by drawing through the section a vertical line and 
finding the distance intercepted on it by the two strings corre- 
sponding to the resultant mentioned ; the product of this inter- 
cept by the pole distance is the required moment. Hence, if 
oe (Fig. 45) is taken as axis of abscissas, the broken line made 
up of oa, ob, oc, od, is a "curve of bending moments." 

119. Design of Beams. — The principles involved in the 
design of beams will not be here fully discussed. Every prob- 
lem in design involves the determination of shears and bend- 
ing moments throughout the beam ; and the graphic methods 
of determining these will alone be considered in the following 
articles. 



$ 2. Beam Siistainiiis; Fixed Loads. 

120. Shear Curve for Beam Supported at Ends. — Let MN 

(Fig. 45) represent a beam supported at the ends and sustain- 
ing loads as shown. Draw the force polygon ABCD, and with 
pole O draw the funicular polygon. The closing line is oe 



BEAM SUSTAINING FIXED LOADS. gg 

(marked also M"N"), and OE drawn parallel to M"N" fixes B, 
thus determining DE, EA, the reactions at the supports. 

Take M'N' as the axis of abscissas for the shear curve. 

The shear at every section can be at once taken from the 
force polygon. For, remembering the definition of external 
shear (Art. 115) we have : 

Shear at any section between M and P is EA (positive). 

Shear at any section between P and Q is EB (positive). 

Shear at any section between Q and R is EC (negative). 

Shear at any section between R and N is ED (negative). 

Hence the shear curve is the broken line drawn in the 
figure. 

121. Moment Curve for Beam Supported at Ends. — As in 
Art. 118, it is seen that the funicular polygon already drawn 
(Fig. 45) is a bending moment curve for the given forces. For 
the bending moment at any section of the beam is equal to the 
corresponding ordinate of this polygon, multiplied by the pole 
distance. 

For simplicity, it will be well always to choose the pole so 
that the pole distance represents some simple number of force- 
units. 

The sign of the bending moment is readily seen to be nega- 
tive everywhere, according to the convention already adopted 
(Art. 47). 

122. Shear and Moment Curves for Overhanging Beam. — 

Consider a beam such as shown in Fig. 46, supported at Q and 
T, and sustaining loads at M, P, R, S, and N. 

This case may be treated just as the preceding, care being 
exercised to take all the external forces (loads and reactions) in 
order around the beam. 

The construction is shown in Fig. 46. First, the reactions at 
Q and T are found as in the .preceding case, by drawing the 
funicular polygon, finding the closing line, and drawing OG 
parallel to it. The reactions are FG and GA. Then the value 



lOO 



GRAPHIC STATICS. 



of the external shear can be found for any section from the 
definition, and is always given in magnitude and sign by a cer- 
tain portion of the force polygon ABCDEFGA. The resulting 
curve is shown in the figure, M^N' being the line of reference. 







Fig. 46 . 

Similarly, from the definition of bending moment, and the 
principle of Art. 56, the bending moment at any section is 
equal to the ordinate of the funicular polygon multiplied by the 
pole distance. The polygon is shaded in the figure to indicate 
the ordinates in question. It is seen that the bending moment 
is positive at every section of the beam. 

123. Distributed Loads. — In all the cases thus far discussed,, 
the loads applied to the beam have been considered as concen- 
trated at a finite number of points. That is, it has been 
assumed that a finite load is applied to the beam at a point. 
Such a condition cannot strictly be realized, every load being 
in fact distributed along a small length of the beam. If the 
length along which a load is distributed is very small, no im- 
portant error results from considering it as applied at a point. 

In certain cases a beam may have to sustain a load which is 
distributed over a considerable part of its length, or over the 
whole length. Such a load may be treated graphically with 
sufficient correctness by dividing the length into parts, and 



BEAM SUSTAINING MOVING LOADS. iqi 

assuming the whole load on any part to be concentrated at a 
point. The smaller these parts, the more nearly correct will 
the results be. 

It may be remarked by way of comparison that while algebraic 
methods are most readily applicable to the case of distributed 
loads, the reverse is true of graphic methods, which are most 
easily applied in the very cases in which analytic methods 
become most perplexing. 

124. Design of Beam with Fixed Loads. — The above exam- 
ples are sufficient to explain the method of treating any simple 
beam under fixed loads. Under such loading the shear and 
bending moment at each section of the beam remain unchanged 
in value, and no further discussion is necessary as a preliminary 
to the design of the beam. We proceed next to the case of 
beams with moving loads. 

§ 3. Beam Sustaining Moving Loads. 

125. Curves of Maximum Shear and Moment. — When a beam 
sustains moving loads, the shear and moment at any section do 
not remain constant for all positions of the loads. In such a 
■case it is the greatest shear or moment in each section that is 
to be used in designing the beam. 

A ciLi've of maxiimim shear is a curve of which the ordinate 
at each point represents the greatest possible value of the shear 
at the corresponding section of the beam for any position of the 
loads. 

A cw've of inaxiniiim moment is a curve whose ordinate at 
■each point represents the greatest possible value of the bending 
moment at the corresponding section of the beam for any posi- 
tion of the loads. 

In the following articles will be explained a method of deter- 
mining any number of points of the curves just defined, in the 
case of a simple beam supported at the ends. The moving 



102 GRAPHIC STATICS. 

load will be taken to consist of a series of concentrated loads 
with lines of actions at fixed distances apart. An example of 
such a load is the weight of a locomotive and train ; the 
points of application of the loads being always under the several 
wheels. 

On PI. Ill is represented a load-series., consisting of two loco- 
motives followed by a train w];iose weight is assumed to be 
uniformly distributed. The numbers given represent half the 
total weight of the train, being the weight borne by each of the 
two beams or trusses which sustain a single-track railroad. 

126. Position of Loads causing Greatest Shear at a Given 
Section. — In order to determine' the position of a given series 
of loads which causes the greatest positive shear at a given 
section of a beam, consider first the way in which the shear due 
to a single load varies as the position of the load changes. 

Let ^1^1 (PI. Ill) represent the beam, and F-^ the section 
under consideration. A load in any^position between F-^ and B^ 
causes at F-^ a positive shear equal in magnitude to the reaction 
at A^; a load between A■^^ and F^ causes at F^ a negative shear 
equal to the reaction at B-^. 

Let A^B^ = /, A^F^ = /p F^B^ = 4. A load P on A^F^, dis- 
tant X from A-j^, causes at F^ a shear j-; as the load moves 

p/ 
from A-j^to Fi^ this shear varies from o to — — J. A load P on 

Pr 
F^B-^, distant x from B^^, causes at F^ a shear — ^; as the load 

PI 
moves from B^ to F-^ this shear varies from o to — ?. These 

results may be represented graphically as follows : 

htfliLence line. — From a given reference line A^B^ (PL III 
{C)\ let ordinates be erected such that the ordinate at any point 
J^ represents (to a convenient scale) the shear' at F^ due to a 
unit load aty^. The line joining the extremities of these ordi- 
nates is called an "influence line " for the shear in question. 



BEAM SUSTAINING MOVING LOADS. 



103 



The above values of the shear due to a load P in any position 
show that for the portion of the beam A-J^-^ the influence line is 
a straight line A^F^, P^i^i being equal in magnitude to -i, and 
being laid off negatively ; while for the portion F-J3-^ the influ- 
ence line is the straight line B^F^,, ^2^?, being equal to -|, and 

being laid off positively. The two lines A^F^ and F^^B^ are seen 
to be parallel. 

The influence line shows at a glance the effect of a load in 
any position in producing shear at /^j. Thus, a unit load at J-^ 
causes a positive shear equal to J^J^i ^-^^d a load P at J^ causes 
a positive shear equal to P ^J<2,Jz- 

The resultant shear at F^ due to any number of loads may be 
found by multiplying each load by the corresponding ordinate 
of the influence line and taking the algebraic sum of the 
products. 

Consider, now, the effect of any series of loads brought on 
the beam from the right. Every load causes positive shear at 
i^i until the foremost load reaches the section, and the effect of 
each load increases as it approaches F-^. As soon as any load 
has passed the section, its effect is to cause negative shear, this 
effect decreasing as the load recedes from the section. 

It thus appears that, in order that the shear at a given section 
may have its greatest positive value, {a) there should be little or 
no load to the left of the section, {b) the portion of the beam to 
the right of the section should be fully loaded, and (<;) the loads 
at the right should be as near the section as possible. 

These rules are not sufficiently definite to serve as an exact 
guide. Suppose the first load to be just at the right of F-^, and 
consider the effect of advancing the whole series of loads. As 
the first load passes the section the shear diminishes by an 
amount equal to that load. But as the loads advance the effect 
of every load in producing reaction at A-^ increases (thus increas- 
ing the shear at F^), while no further decrease occurs until the 
second load passes the section. The net result of bringing the 



104 



GRAPHIC STATICS. 



second load up to the section may or may not be to increase 
the shear over the value it has when the first load is just at 
the right of F-^. It is thus uncertain, without computation, 
whether the first load or some succeeding load should be brought 
up to the section in order to cause the greatest positive shear. 

Instead of resorting to trials to determine for which position 
the shear is greatest, we may apply a simple rule which will 
now be deduced. 

Let P^ = magnitude of foremost load, Pg = magnitude of sec- 
ond load, etc., W being the total load on the beam. Let / = 
total span, and m = distance between P^ and P^. Let x = dis- 
tance from B-^ to the center of gravity of W^when P^ is at the 
given section. Let us compute the shear when P^ is just at 
the right of the section, and then determine the effect of moving 
all loads to the left, until the second load comes to the section. 

When P^ is at the right of the section, the shear is equal to 
the reaction at A^ say R. Then (calling V the external shear) 
we have 

V=P = ^. 

If now the loads be moved until P^ comes to a point just 
at the right of the section, the reaction due to W becomes 

W- , and the shear at the section becomes 

The increase of the shear is, therefore, 

W {x+ ni) p Wx Wm ^ 

This increase is plus or minus according as -— — is greater or 

P I 
less than P-, ; that is, as Wis greater or less than — L. Hence 

m 



the following rule 



BEAM SUSTAINING MOVING LOADS. 



105 



The maximum positive shear in any section of the beam occurs 

■when the foremost load is irifinitely near the section, provided W 

PI ■ PI 

is not greater than — ^ . If W is greater tJian — 5^, tJie greatest 
m ni 

shear will occur when some succeeding load is at the section. 

In the above discussion it has been assumed that in bringing 
P^ up to the section no additional loads are brought upon the 
beam. If this assumption is not true, let W be the new load 
brought on the beam, and x' the distance of its center of 
gravity from the right support when P^ is at the section. Then 
the shear corresponding to this position is 

and the increase of shear due to the assumed change in the 
position of the load is 

Wm , W'x' o 

Hence, in the statement of the above rule, we have only to 
substitute Wm+W'x' for PVm ; or, W+!^^—^ for W. The 

additional load W may be neglected except when the condition 

P I 

W= — — is nearly satisfied ; for the term W'x' will always be 
m 

small. 

The application of this rule is very easy, and will save much 
labor in the graphic construction of the shear curve. 

If it is found that the first load should be past the section 
for the position of greatest shear, we may determine whether 
the second or the third should be at the section by an exactly 
similar method.. We have only to apply the above rule, substi- 
tuting /*2 for -^i> ^nd for m the distance between P^ and Pg. 

127. Determination of Greatest Shear. — Having determined 
what position of the load-series causes the greatest shear at a 
given section, the value of this shear may be determined by 



I06 GRAPHIC STATICS. 

the method already explained in the discussion of fixed loads 
(Art. 120). 

Draw in succession the lines of action of the loads at their 
proper distances apart (1-2, 2-3, . . ., PI. III). Draw the force 
polygon * 1-2-3- • • •> choose a pole O, and construct a funicular 
polygon for the series of loads. The same lines of action may 
be used for all positions of the load-series, for instead of moving 
the loads in the desired direction, the loads may be regarded as 
fixed and the beam moved in the opposite direction. 

At (5), PI. Ill, is shown a beam A-^B-^, of 64 ft. span, in such 

position that the shear at the section F^ 16 ft. from A^, has its 

greatest possible value. The second load is at the section 

instead of the first, for applying the test we have P^ = d,ooo lbs., 

P I 
1=64. ft., m — S.i ft., —i-= 63200 lbs. ; while the total load on 
in 

the beam is 104,000 lbs. when the first load is at the section. 

The beam being in the position shown at (5), vertical lines 
drawn through A-^ and B-^ intersecting the strings 01 and 02' at 
A and B, determine AB, the closing string of the funicular 
polygon for the system of forces acting upon the beam in the 
assumed position. If OK is the ray drawn parallel to AB, the 
reactions at A^ and B^ are Ki and 2' K respectively. The load 
2-3 being just at the right of the section F^, the shear at that 
section is equal to the reaction at A-^^ minus the load 1-2; its 
value is therefore K2 in the force polygon. 

The diagram of maximum positive shear is shown at {A), 
PI. Ill, the ordinates representing greatest shear due to the 
moving loads being laid off upward from A-^B-^. Thus the value 
just determined is laid off from the point F-^. The entire curve 
is shown at {A\ but the construction is given only for the 
section F-^. 

The foregoing construction has referred only to moving loads. 
The actual greatest shear at any section is found by combining 

* The lines representing the loads are for convenience drawn upward instead of down- 
ward. This does not, of course, affect the construction. 



BEAM SUSTAINING MOVING LOADS. 



107 



the maximum live-load shear with the shear due to permanent 
loads. Shears due to the dead loads are represented at (A), 
PI. Ill, being determined as follows. 

128. Shear Curve for Combined Fixed and Moving Loads. — 

Let the beam sustain a fixed load of 25000 lbs. uniformly dis- 
tributed along the beam. The shear close to the left support 
due to this load is equal to the reaction, or 12500 lbs. ; and 
decreases as we pass to the right by ^^-^-^ lbs. for each foot. 
At the middle of the beam the shear is zero, and at the right 
support it is — 12500 lbs. Hence the shear curve is a straight 
line, and may be drawn as follows : From A■^^ lay off an ordinate 
downward representing 12500 lbs., and from B^ an ordinate 
upward representing 12500 lbs.; the straight line joining the 
extremities of these ordinates is the shear curve for the fixed 
load. Positive shears are laid off downward and negative 
shears upward for the reason that, if this be done, the greatest 
positive shear at any point due to fixed and moving loads is 
represented by the total ordinate measured between the shear 
curves for fixed loads and for moving loads. It is seen that 
at a certain point somewhere to the right of the center of the 
beam this greatest shear is zero, and for all sections to the 
right of this point it is negative. This point is determined by 
the intersection of the two shear curves. 

129. Approximate Position of Loads causing Greatest Bending 
Moment at a Given Section. — Let it be required to determine the 
greatest bending moment at the section F^ of the beam A-^B^, 
shown on PI. III. Let A^F^=^/-^^, F^B^=^l^, A^B^ = l^-\-l^^l. 
Consider the effect of a single load in any position. 

The bending moment at F^ due to a loadP on A-^F-^, distant x 

PI X 
from A^, is — |— . For a load P on F^B^, distant x from B^^, the 

. PI X 
bending moment at F^ is — j—. The variation of each of these 

values as the position of the load changes may be represented 
graphically as follows : 



I08 GRAPHIC STATICS. 

Influence line. — From A^B^ ((-^)> PI- III)j ^^t ordinates be 
erected such that the length of the ordinate at any point repre- 
sents, on a convenient scale, the bending moment at F-^, due to 
a unit load at the corresponding point of the beam. The line 
joining the extremities of these ordinates is the "influence line" 

for bending moments at F^. 

I X 

At a distance x from A^, the ordinate has the value -^, which 

// 
varies from o at A^_^ to ~ at F^. At a distance x from B^, the 

I X . . II 

value of the ordinate is -^, which varies from o at B^ to -^ 

at /^2- The influence line therefore consists of the two straight 

portions A^F^, -^3^2' where F^F^ = ^^- The ordinates are laid 

off downward, the bending moment being always negative 
according to the convention already adopted (Art. 47 and 
Art. 116). 

The influence line having been drawn, the bending moment 
at F-^ due to a load at any point J^ may be found by multiply- 
ing the load by the corresponding ordinate J^J^- The bending 
moment at F-^ due to the combined action of several loads is 
found by multiplying each load by the corresponding ordinate 
of the influence line and adding the products. 

Since the sign of the bending moment due to a load is the 
same, whatever its position, and since the bending moment due 
to a given load increases as the load approaches the section, the 
following general principle may be stated : 

The bending moment at any section has its greatest value 
when the beam is as fully loaded as possible, with the heaviest 
loads near the section. 

This principle serves only as a rough general guide. An 
exact rule will be deduced in the following Articles. 

130. Load Polygon. — Let a curve or broken line be drawn, 
of which the abscissa is horizontal, and the ordinate at any 
point represents (on the assumed force scale) the total load up 



BEAM SUSTAINING MOVING LOADS. 109 

to that point, measured from some fixed point in the load series 
Such a curve or broken line may be called a load polygon. In 
the figure on PI. IV, Q! R' is a load polygon for the series of 
train loads shown on PI. Ill, QR being a funicular polygon 
for the same series. 

A simple relation exists between the load polygon and the 
funicular polygon, which is of use in the discussion of the prob- 
lem now before us. 

Let X and J be the coordinates of any point of the funicular 
polygon, the /-axis {L F) being vertical, and the ,f-axis any hori- 
zontal line LX. Let x' , y' be the coordinates of the correspond- 
ing point of the load polygon, referred to a vertical axis {L' Y'), 
and a horizontal axis (L'X') passing through the pole of the 
force polygon used in the construction of the funicular polygon. 

Consider any point of the funicular polygon, for example a 

point on the string 02'. The value of — at this point is the 

dx 

tangent of the angle between 02' and the horizontal. But since 
02' is parallel to the corresponding ray of the force polygon, 

A/' 

this tangent is equal to —, where H is the pole distance used 
in drawing the funicular polygon. That is, 

dy _ y' 
d^~H' 

or y=Nf- (r) 

ax 

131. Variation of Bending Moment at Any Section of a Beam. 

— Let ^ii?i (at (A), PI. IV) represent one position of a beam 
of span /, and let F^ be any section at which it is desired to 
study the variation of the bending moment. Let l-^ = A-^F-^, 
l^ = F^B^, and ^■ = horizontal distance of A^ from the vertical 
axis V v. Through A-^^, B^, and F^ draw vertical lines, inter- 
secting the funicular polygon in A, B, F, and the load polygon 
in A', B', F'. Let the vertical through F-^ intersect AB in G, 
and A'B' in G'. Let the ordinates of A, B, F, G, measured 



no GRAPHIC STATICS. 

from LX, be denoted by a, b, f, g; and the ordinates of A' , B\ 

F', G' , measured from L'X' , by a' , V ,f\ g'. 

If M denotes the bending moment at F^, and H the pole 

distance, 

M=NxFG=II(g-/); (2) 

and it will now be shown that the rate of change of this bend- 
ing moment {i.e. the change per unit of horizontal displacement) 
as the beam moves horizontally while the loads remain sta- 
tionary is equal to F'G' or g-'—f. 

AG I 
Since AGB is a straight line, and -^^jr=^, it follows that 

b-g K 

or g=ja-hjd; 

FG=g-fJja + ^jb-f. (3) 

In an exactly similar manner it may be shown that 

F'G'^g^-f'Jja'-h^f-f (4) 

From (2) and (3), 

- M=II(g-f) = H(^ja + ^jd-fj (5) 

If the beam moves horizontally, the abscissas of A^, B^ and F^ 
change at the same rate. Differentiating (5) with respect to ^, 



dz 



_ Tj-fk da /j db df\ 
\l dz I dz dz J 



.^ da db , df . , , r dy ^ . j^ 

Now —r, —rt and ~ are equal to the values of -— at A, B, 
dz dz dz dx 

and F, respectively; therefore, from equation (i) of Art. 130, 

a'=H—, b' = H—,f' = H^; 
dz dz dz 



BEAM SUSTAINING MOVING LOADS. m 

hence ^— =^^'+4'^'— /'; 

as I I 

or, comparing with equation (4), 

^=g'-f'^F'G' (6) 

132. Condition for Maximum or Minimum Bending Moment at 
Any Section. — As the beam moves relatively to the loads, FG 
(proportional to M) in general varies. As it passes through a 

maximum or a minimum value, must equal zero ; but, from 

ds 

the relation just proved, this makes F'G' = 0. That is. 

When the bending moment at any given section of the beam is 
a maximum or a minimum, tJie line A'B' intersects tJie load 
polygon at a point directly opposite the section considered. 

This condition may be stated in algebraic form as follows : 
Let W-^ denote the total load between A^ and F^^ W^ the 

load between F^ and B^ and W the total load on the beam. 

Evidently, 

Wi=/'-a'; W^ = b'-f] W=b'-a'. 

If the points F' and G' coincide,/' — «' and b' —f are propor- 
tional to A' G' and G'B' , that is to /j and 4, and therefore, 

U\_W^_W^ 

t-l ley i' 

That is, when the bending moment at any section of the beam 
is a maximum or a minimum, the average load per unit length 
on each segment is equal to the average load per unit length on 
the whole span.* 

* It is to be noted that if, at any point, there is a load which is concentrated in the strict 

mathematical sense (i.e., a finite load applied at a mathematical point), the value of -^ 

dx 
suffers a sudden change of value at the point of application of that load, and the equation 

does not hold at that point. This consideration does not, however, invalidate the above 
conclusions. For {a) a. concentrated load in the strict mathematical sense does not occur 



112 GRAPHIC STATICS. 

133. Discrimination between Maxima and Minima. — The 

magnitude of the bending moment is 

M=H{g-f)- 

it is obvious from the form of the funicular polygon that g is 
always greater than f. 

li g' —f is positive, ^— _/" (and therefore M) increases as s in- 
creases {i.e., as the beam is displaced in the positive direction). 
If g' —f is negative, M decreases as z increases. Therefore, 
as M passes through a maximum value, the sign of g' —f 
changes from plus to minus ; and as M passes through a mini- 
mum value, g' —f changes from minus to plus (the displace- 
ment of the beam being in the positive direction, that is, toward 
the right as the figure is drawn on PI. IV). 

If the load series consists of concentrated loads, g' —f will in 
general pass through zero only when a load is at one of the 
three points A-^, B^, F^. By applying the test just given, it may 
be seen that a maximum value can result only when a load 
passes F^. 

In applying the criterion for a position of maximum or mini- 
mum value of the bending moment, and in distinguishing posi- 
tions causing maximum stress from those causing minimum 
values, it is not necessary to draw the line A'B' for every trial 
position of the beam. A movable strip of paper representing 
the beam may be laid in any desired position, the points A' and B^ 
approximately located by inspection, and a thread stretched 
between them. If the thread crosses the load polygon at a 
point directly opposite the section in question, the condition for 

in practice, and {d) the ideal case of a finite load applied at a mathematical point may be 
treated as the limit of a case of distributed loading, so that even in that case the conclu- 
sions above derived hold, with slight change in the form of statement. 

The series of concentrated loads actually coming upon a bridge in practice gives a load 
polygon which, while approaching the form shown on PI. IV, differs from it in that the 
portions under the loads are not strictly vertical lines, but are lines of variable (very steep) 
inclination. 

In applying the condition just deduced, a load at either of the three points A^, B-y, F^, 
is to be regarded as distributed in an arbitrary way along the small element of horizontal 
distance near the line used as its line of action. 



BEAM SUSTAINING MOVING LOADS. 



113 



maximum or minimum is fulfilled. Only positions which bring 
a load at the section need be tried. The accurate location of 
A' and B' is usually unnecessary.* 

At {B), PI. IV, is shown the beam A^B^ in such a position 
thsit F'G' = 0. It is seen that this position corresponds to a 
maximum value of the bending moment at F^^ ; for a displace- 
ment of the beam toward the right (positive) makes £■' —f 
negative, while a displacement toward the left makes g' —f 
positive. 

134. Determination of Bending Moment. — Having determined 
the position of the load series causing greatest bending moment 
at a given section, the value of the bending moment may be 
computed by the method already explained in the case of fixed 
loads. At {M), PI. Ill, is represented the position of the beam 
A-^B-^, which causes the greatest bending moment at the section 
F-^, distant 16 ft. from A-^, the span being 64 ft. That the con- 
dition for a maximum is satisfied in this position may easily be 
verified. The total load on the beam is 1 12000 lbs., one-fourth 
of which is 28000 lbs. By regarding 5000 lbs. of the load 3-4 
as belonging to the segment A-^F-^, the load on this segment is 
28000 lbs. It is seen also that the beam is fully loaded, and 
that the heaviest loads are near the section. Moreover, the 
condition for maximum is not satisfied for any other position 
near the one chosen. 

The closing string of the funicular polygon for all external 
forces acting on the beam in the position {M) is AB. The 
bending moment at F-^ is the product of the ordinate FG by the 
pole distance. 

At {B\ PL III, is shown the diagram of maximum bending 
moments for all sections of the beam. From A-^B-^ are laid off 
downward ordinates proportional to the maximum bending 
moments. The length FG just determined is laid off from F^; 
this must be multiplied by the pole distance (i 00000 lbs. in the 

* See paper by Professor Henry T. Eddy, Trans. Am. Soc. C. E., Vol. XXII, 



114 GRAPHIC STATICS. 

diagram). The complete curve is drawn, but the construction 
for determining values of the maximum bending moment is 
shown only for the section F-^. 

135. Moment Curve for Fixed Loads. — The greatest bending 
moment due to moving loads must be combined with the bend- 
ing moment due to fixed loads. If the fixed load is uniformly- 
distributed, as already assumed in the computation of shear, 
it may be divided into parts, each assumed concentrated at 
its center of gravity, and a funicular polygon drawn, using the 
same pole distance employed in the force diagram for moving 
loads. The ordinates of this funicular polygon may be laid off 
■upward from the line A-J3^, and their ends joined to form a 
continuous curve. The total ordinate from this curve to that 
already drawn for moving loads represents the true greatest 
bending moment at the corresponding section of the beam. 
The curve is shown at {B), PL III, but the construction is 
omitted, since it involves no new principle. 

It is to be remembered that the bending moment found for 
any section is a possible value for the other section equally 
distant from the center of the beam, since the train may be 
headed in the opposite direction, and the same construction 
made, viewing the beam from the other side. The same state- 
ment holds as to shears. 

136. Design of Beam sustaining Moving Loads. — In designing 
a beam to sustain moving loads, the greatest shear and bending 
moment that can come upon it for any position of the loads must 
be known for every section. The methods above given are 
sufficient for the determination of these quantities; and the 
problem of designing the beam will not be here further discussed. 



CHAPTER VII. TRUSSES SUSTAINING MOVING 

LOADS. 

§ I. Bridge Loads. 

137. General Statement. — The most important class of trusses 
sustaining moving loads is that of bridge trusses. The two 
main classes of bridges are highway bridges and railway bridges. 
The forms of trusses most commonly used differ for the two 
classes, as do also the amount and distribution of loads. 

Before the design can be correctly made, the weights of the 
trusses themselves must be known. Since these weights de- 
pend upon the dimensions of the truss members, they cannot 
be known with certainty until the design is completed. The 
remarks made in Art. 82 regarding the design of roof trusses 
are here applicable. 

In the following articles we shall give data available for 
preliminary estimates of truss weights. 

Railway bridges of short span are frequently supported by 
rolled or built beams. When the span is longer than 100 ft. 
trusses should be used. (Cooper's " Specifications for Iron and 
Steel Railroad Bridges.") 

138. Loads on Highway Bridges. — (i) Permanent load. — 
The permanent load sustained by a highway bridge truss in- 
cludes the weight o.f the truss itself, of the lateral or "sway" 
l)racing, of the floor and the beams and stringers supporting it. 
These weights are all subject to much variation, but, for pur- 
poses of preliminary design, the following formula, taken from 
Merriman's " Roofs and Bridges," may be used. 

"5 



Il6 GRAPHIC STATICS. 

Let w = weight of bridge in pounds per linear foot; /=span 
in feet ; <5 = width in feet. Then 

W= 140+12 (5 + 0.2 bl—0.4. I. 

(2) Snow load. — The weight of snow may be taken as in 
case of roof trusses (Art. 84). The values there given are 
probably in excess of those ordinarily employed in practice. 

(3) Wind load. — The pressure of wind striking the bridge 
laterally is resisted by the chord members together with the 
lateral bracing. These constitute horizontal trusses, in which 
the stresses are to be found in the same way as for the main 
trusses of the bridge. As the determination of wind stresses 
requires the use of no special methods or principles, they will 
not be here considered. The student is referred to Burr's 
" Stresses in Roofs and Bridges," Merriman's " Roofs and 
Bridges," and other available works for a complete discussion 
of wind pressure and its effects on bridge trusses. 

(4) Moving load. — The most dangerous moving load for a 
highway bridge is usually a crowd of people or a drove of 
animals. This is commonly taken as a uniformly distributed 
load, which may cover the whole bridge or any portion of it. 
Its value is variously taken at from 60 lbs. to 100 lbs. per square 
foot of area of floor, depending upon the span and upon local 
conditions. 

It may be that in certain cases the greatest stresses will 
result from the passage of heavy pieces of machinery over the 
bridge, as, for example, a steam road roller. This should of 
course be considered in the design. 

For a complete discussion of loads on highway bridges, the 
student is referred to Waddell's " Highway Bridges." 

139. Loads on Railway Bridges. — (i) Permanent load. — 
The permanent load on a railway bridge includes {a) the weight 
of the track system, which is known or may be determined at 
the outset ; ib) the weight of longitudinal stringers and cross- 



BRIDGE LOADS. I17 

beams, which can be determined before the trusses are designed ; 
and (c) the weights of trusses. The weight of the track system 
may be taken at 400 lbs. per linear foot for a single track. 
(See Burr's " Stresses in Bridge and Roof Trusses " ; Cooper's 
" Specifications for Iron and Steel Railroad Bridges " ; Merri- 
man's " Roofs and Bridges.") The total weight of track system 
and supporting beams and stringers varies from 450 lbs. to 
600 lbs. per linear foot. (Merriman.) For spans less than 100 
feet, Merriman gives the following formulas, in which zv is the 
total dead load of the bridge in pounds per linear foot, and / is 
the span in feet : 

For single track, w= 560+ 5.6 /. 

For double track, w= 1070+ 10.7 /. 
See also Art. 8 of Burr's work above cited. 

(2) Snozv and zvind. — Railway bridges usually offer little 
opportunity for the accumulation of snow. Wind pressure is, 
however, an important factor. Besides the pressure upon the 
bridge itself, the pressure upon trains crossing the bridge must 
be considered. The latter is a moving load and may be dealt 
with in the same way as other moving loads. Its amount may 
be computed from the area of the exposed surface of the train 
and the known (or assumed) greatest pressure due to wind 
striking a vertical surface (Art. 85). 

For further discussion of wind pressure, the student is referred 
to the works already cited. The graphic methods of deter- 
mining stresses due to wind will be evident when the methods 
for vertical loads given in the following articles are understood. 

(3) Moving loads. — The moving load to be supported by a 
bridge consists of the weights of trains. Such a load is applied 
to the track at a series of points, namely, the points of contact 
of the wheels. But the load is applied to the trusses only at 
the points at which the floor beams are supported. Hence the 
actual distribution of loads upon the truss is somewhat com- 
plex. It was formerly common to substitute for the actual 



Il8 GRAPHIC STATICS. 

load a uniformly distributed load, thus simplifying the problem) 
of determining stresses. It is now more usual to consider the 
actual distribution of loads for some standard type of locomo- 
tive used by the railroad concerned, or specified by its engi- 
neers. For examples of such distributions the student is 
referred to Cooper's " Specifications " already cited ; also to^ 
PL III, and to the following portions of this chapter. 

140. Through and Deck Bridges. —A bridge is called through 
or deck according as the floor system is supported at points of 
the lower or of the upper chord. In the former case, if the 
trusses are too low to require lateral bracing above, they are 
called /^;2j trusses. 

The weight of the truss itself is to be divided between the 
upper and lower joints. But the weight of the floor system 
and of the supporting beams and stringers comes wholly at the 
lower joints of a through bridge, or at the upper joints of a 
deck bridge. The moving load is, of course, applied at the 
same joints at which the floor system is supported. 

If the floor system is supported directly upon the upper 
chord, as is sometimes the case, the moving load and part of 
the dead load produce bending in the chord members ; the 
design is otherwise unaffected by this construction. 

§ 2. Truss regarded as a Beam. 

141. Classification of Trusses. — Since a bridge truss acts as 
a practically rigid body resting on supports at the ends or other 
points and sustaining vertical loads, it may be regarded as a 
beam, and trusses may be classified in the same way as beams 
(Art. 114). The only class to be here considered is that of 
simple trusses, or such as may be regarded as rigid bodies for 
the purpose of determining the reactions. 

Cantilever trusses and continuous trusses are defined like the 
corresponding classes of beams (Art. 1 14). The most impor- 
tant case is that of a truss simply supported at the ends. 



TRUSS REGARDED AS A BEAM. 



119 




142. External Shear for a Truss. — If a truss be regarded as 
a beam, the external shear, resisting shear, and internal shear- 
ing stress at any section may be defined just as in Art. 115. 
In some forms of truss a knowledge of the external shear at 
any section makes it possible to compute readily the stresses in 
certain truss members. Thus, in the portion of a truss repre- 
sented in Fig. 47, let the section 
MN cut three members, of which 
two are horizontal. (The member 
x^y^ is disregarded.) Since each 
member can exert forces only in 
the direction of its length, the ex- 
ternal shear in the section MN 
must be wholly resisted by the di- 
agonal xy ; and the internal force 
in xy must be such that its resolved 
part in the vertical direction is equal in magnitude to the exter- 
nal shear in the section. Let the external shear V be repre- 
sented by YZ (Fig. 47) ; draw YX parallel to yx and ZX 
horizontal; then JfFwill represent the stress in xy. If V I's, 
positive (Art. 115), the stress in xy is a tension. If Fis nega- 
tive, the stress is a compression. If the member xy were 
replaced by one sloping the other way from the vertical, these 
statements as to kind of stress would be reversed. 

If no two of the three members cut by any section are par- 
allel, the stresses cannot be computed so simply, since all may 
contribute components of force to resist the external shear. 



V 


IN 






Fig. 47 


\ 



143. Bending Moment for a Truss. — In many cases the 
stresses in the truss members can be found from the values of 
the bending moment at different sections. 

Thus, in Fig. 49, let a section MN be taken cutting three 
members as shown, and let the origin of moments be taken at 
the point of intersection of two of them (as bx and xy) ; then 
since the moments of the internal forces in these two will be 



120 GRAPHIC STATICS. 

zero, the resisting moment is equal to the moment of the inter- 
nal force in the third member ajy. The arm of this latter force 
is the perpendicular distance of its action line from the origin. 
Call it k, and let / represent the force itself, and M the bending 
moment. Then, numerically, 

M 
ph = M. .•./ = ^- 

If M is positive (in accordance with the convention of Art. 
47), / must act from right to left, hence the stress in ay is a 
compression. If M is negative, the stress is a tension. (It 
must be remembered that the forces whose moments make up 
the bending moment M act upon the portion of the truss to 
the left of the section.) 

§ 3. Truss sustaining Atiy Series of Moving Loads. 

144. General Method of Determining Maximum Stresses in 
Truss Members. — When a truss carries a definite series of mov- 
ing loads, the problem of determining the maximum stress in 
any member involves (i) the determination of the proper posi- 
tion of the loads and (2) the actual computation of the stress for 
a known position of the loads. The second part of this problem 
involves only the principles already explained in the discussion 
of trusses carrying fixed loads. The first part requires special 
■consideration. 

Although the treatment of trusses with parallel chords is in 
some respects simpler than that of the case of non-parallel 
chords, it is advantageous to consider the problem in as general 
a manner as possible. The following discussion will therefore 
refer to the general case of a truss of any form, subject only to 
the restriction that there are no redundant members, so that the 
truss may be completely divided by cutting three members, of 
which any chosen member may be one.* No restriction will be 

* A case not included in this general description, while still offering a determinate 
problem, is that of a truss with subordinate bracing. This important case will be consid- 
ered in a following discussion. 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 12 1 

made as to the distribution of the loads. Trusses with parallel 
chords, and particular distributions of loading, may be treated 
as special cases of the general problem.* 

145. Computation of Stress when Position of Loads is 
Known. — The general "method of sections" (Art. 69) may be 
applied to determine the stress in any member when the posi- 
tion of the load series is known. 

Thus, referring to PL V, let 1-2, 2-3, . . . represent the lines 
of action of the successive loads, 1-2-3- • • • the force polygon 
for the loads, and let a funicular polygon be drawn as shown. 
The truss is represented in two positions marked (A) and (B) 
respectively. In connection with (A) is shown the construction 
for determining the stress in the member 6\Z)j, and in connec- 
tion with (B) the construction for determining the stress in CiH. 
Referring to the former case, let a section be taken through the 
three members C^D^, C^F^, H^F^ ; then the internal forces in 
the three members cut are in equilibrium with the external 
forces acting on the portion of the truss on either side (as the 
left) of the section. These forces are the reaction' at A-^ and all 
the loads from A^ to C^. Applying the principle of moments, the 
origin may be taken at F^, the point of intersection of the lines 
^j/^2 3.nd H^F^j, thus eliminating the forces acting in these lines. 

The moment of any load is equal to the sum of the moments 
of any components which may replace it. Hence for any load 
between A-^ and 6\ (though actually coming upon the truss at 
the panel joints) the moment may be computed as if it were 
applied to the truss at a point in the vertical line through its 
actual point of application on the floor or track. But in the 
case of a load on the panel C^D-^, only one component comes 
upon the portion of the truss to which the moment equation is 
to be applied ; it is therefore necessary to determine the portion 
of every such load which is actually carried at C-^. If loads 
between C-^ and D^ are carried wholly at the two points C^ and 

* The case of parallel chords, with graphic methods of treatment, receives very full dis- 
cussion in a paper by Professor Henry T. Eddy, Trans. Am. Soc. C. E., Vol. XXII (1890). 



122 GRAPHIC STATICS. 

Z?p the portion carried at each of these points is easily deter- 
mined, as will be shown.* 

From ^j and B^ draw vertical lines intersecting the funicular 
polygon at A and B respectively ; the line AB is the closing 
line of the funicular polygon for the external forces acting upon 
the truss when the load series has the assumed position. On 
the panel C-^D-^ is the load 6'f. Treating C-^D-^ as a simple 
beam, the components replacing this load at C-^ and D^ are found 
(Art. 45) as follows : Draw vertical lines through 6\ and D^ 
intersecting the strings 06' and oy' in C and D ; from the pole O 
draw the ray OK parallel to CD ; the required components of 
6'f are represented in the force polygon by 6' K and Kf. The 
same construction might be made for every panel, and it is seen 
that the series of lines such as CD would form the funicular 
polygon for the loads as actually applied to the truss at the 
several joints. But the construction is here needed only for 
the panel C-^Dy since a load on any other panel comes wholly 
upon one of the two portions into which the truss is divided by 
the section taken. 

The sum of the moments of the external forces acting upon 
the left portion of the truss may be found graphically by the 
method of Art. 56. That is, through the origin of moments a 
line is drawn parallel to the resultant of the forces (a vertical 
line in this case) ; its intercept FG, between the strings AB and 
CD, multiplied by the pole distance, gives the required moment. 
The origin being taken at F^, this moment is equal to the 
moment of the internal force in the member C-^D-^ ; hence that 
force may be found by dividing the moment by the perpendicu- 
lar distance from F^ to C-^D-j^. 

Referring next to {B) and considering the member C^H, the 
stress in this member may be found in a similar manner. The 

* It is to be noticed that the actual distribution of any load among the different joints is 
not determinate by simple methods, if the longitudinal beams supporting the floor system 
are supported at more than two points. It will here be assumed that the portion of such a 
beam between two supports acts as a simple beam. The results of this assumption are 
probably as rehable as could be obtained by a more elaborate discussion. 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 123 

points A, B, C, D, F, G have the same meanings in this con- 
struction as in the preceding, but AB and CD must be produced 
in order to find the intercept FG, because the origin of moments 
is at F-^, beyond the end of the span. The moment of the 
internal force in C-^H is equal to FG multiplied by the pole dis- 
tance; dividing this moment by the perpendicular distance 
from F-^ to C^H gives the magnitude of the required force. 

The kind of stress in any member may be determined by 
inspection of the conditions, as in Art. iii. 

. If the stress in a member is determined by the foregoing 
method, the steps of the process are the same, whatever mem- 
ber be under consideration. These steps are as follows : 

(i) A section is taken completely dividing the truss and cut- 
ting only three members, one of which is the member whose 
stress is to be determined, and one a member of the chord car- 
rying moving loads. 

(2) Taking origin at the intersection of those two of the 
three members cut whose stresses are not in consideration, 
equate the sum of the moments of the external forces acting 
upon one portion of the truss to the moment of the internal 
force in the member considered, 

146. Definitions and Notation. — The following definitions 
and notation will conduce to clearness and generality in the ensu- 
ing discussion. 

Let the origin of moments, assumed in determining the stress 
in any member by the foregoing general method, be called the 
vioment-center for that member. 

Let the stress-moment for any member be defined as the mo- 
ment of the stress in that member with respect to the moment- 
center just described. Let the sign of the stress-moment be 
specified in the following manner : 

Rotation being called positive or negative in accordance with 
the convention adopted in Art. 47, let the stress-moment be 
called positive when it resists a tendency of the left portion of 



124 



GRAPHIC STATICS. 



the truss to rotate negatively (and of the right portion to rotate 
positively). 

In all cases let the left and right ends of the span be marked 
A-j^ and B^, and let 6\ and D^ designate the left and right ends 
of the panel cut by the assumed section. Let the vertical line 
through the moment-center cut A-^^B^ (produced if necessary) in 
a point marked F^. 

Distances from left to right being positive, and from right to 
left negative, let the following notation be used : 

^1-^1=^1' ^1^1=4' .^1^1 = ^1 +4 =^; 
CiF^ = «i, F^B>^ = «2, C^D^ = /Zj + «2 = n. 

The sign of each of the four quantities l^, 4, n^, n^ is to conform 
in all cases to the direction, as indicated by the order of the letters 
used in defining it. Thus l^ = A-^F-^ is positive or negative, 
according as F-^ is to the right or to the left of A-^. 

147. Effect of a Single Load. — Inflitence line. — A general 
idea of the position of a load series which will cause the greatest 
stress in any member of a truss may be obtained by considering 
the effect of a single load placed in any position. The details 
of the discussion will vary with the form of the truss, but the 
method of reasoning may be sufficiently illustrated by reference 
to the form shown at {A) or (^), PL V. 

The relative effect of loads in different positions in producing 
stress in a given member may be clearly represented by means 
of an "influence diagram," analogous to the diagrams used in 
the discussion of shear and bending moment in a beam (Arts. 
126 and 129). 

From a given base line let ordinates be erected such that the 
ordinate at any point represents (to a convenient scale) the stress 
due to a unit load at the corresponding point of the span. The 
line joining the extremities of these ordinates is the influence 
line * for the member in question. 

* For a discussion of influence lines, see paper by Professor George F. Swain, Trans. 
Am. Soc. C. E., Vol. XVII, p. 21 (1887). 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 



125 



It will be shown that there is a simple method, of complete 
generality, by which the influence line may be drawn for any 
member whatever. In order to explain this method clearly, it 
is necessary to adopt a convention as to signs. 

In laying off ordinates of the influence line, let positive values 
be drawn upward and negative values downward from the refer- 
ence line. Let the sign of the ordinate agree in all cases zvith 
that of the stress-moment (Art. 146). With this convention, a 
positive ordinate may correspond to either a compressive or a 
tensile stress, depending upon the relation of the moment-center 
to the member under consideration. From the sign of the stress- 
moment the kind of stress may always be readily determined by 
inspection of the truss diagram. 

In Fig. 48 (PI. VIII) are shown the forms assumed by the 
influence line for three typical members of the truss shown ; 
{A) represents the case of the chord member C-JJy (B) that of 
the chord member C-lD{, and (6") that of the web member 
C-^Dy The method of constructing these diagrams will now 
be explained. 

Member of loaded chord. — Consider first the member C-J)-^, 
The moment-center is €{, and F^ falls between Q and D-^, the 
corresponding position of F^ being shown at (yA). Here /j, l^_^, 

7?i, n^ are all positive (Art. 146). A load P between D-^ and B-^, 

Px 
distant x from B^, causes a reaction — ^ at A-^, and a stress- 

Pl X ... 

moment equal to — 1-. This varies directly as x, being o at B^ 

PI [l -.■ji\ ^ 
and — ^-^^-? ^ at D-^. Similarly, the stress-moment due to a 

. PI x 

load P between A^ and 6\, distant x from A^, is — ?— , which 

PI (I —n") 
varies from o at A-^ to — ^^^ i^ at C^. If the moment-arm 

of the required stress is /, the ordinates of the influence line at 
Q and Dc^ are 

and the straight lines A^C^ and D^B^ are the portions of the 



126 GRAPHIC STATICS. 

influence line for the corresponding portions of the span. A 
load P between C-^ and D-^, distant x from C-^, comes upon the 

truss partly at 6\ and partly at D^ ; the former part is — ^^ K 

p ^^ 

the latter — . The total stress-moment due to such a load is 

It 
therefore 

P{n-x) l^{l^-n^ ^ Px l^{l^-n^ ^ 

n I 11 I ^ 

putting P=i and dividing by t, this expression gives the ordi- 
nate of the influence line for any point between C-^ and D-^. The 
influence line for this portion of the span is therefore the straight 
line C^D^. It will be noticed that, if the straight line B<^D^ be 

continued, its ordinate at F,^ will be F^f^ = ^^\ if A^C^ be con- 
tinued, its ordinate at /^2 will have the same value. 

Member of chord not carrymg moving loads. — Next consider 
the member C-^D-^. The moment-center is D-^, F^ falls at D^, 
and F^ coincides with B^ as shown at (F), Fig. 48. If the 
reasoning used in the preceding case be repeated, it is seen that 
the values there given for the ordinates of the influence line 
apply also to this case. Since in the present case n^ = o, the 

that A^CoD^, is a straight line. 

Web member. — For the member C{D-^ the moment-center is 
at the intersection of C^D^ and C-J)^ ; F-^ falls to the left of A^, 
the corresponding position of F^ being shown at (yC\ Fig. 48. 
Here it will be noticed that /^ and 71-^ are negative. If, however, 
the method used in the preceding cases is again applied, it is 
found that the values above given for the stress-moment due to 
loads on D-J^-^ and A-^C^ are correct for the present case both in 

magnitude and in sign. Thus, for a load P on D-J^-^, distant 

Px 
X from j5*j, the reaction at ^j is — — , and the stress-moment is 

Pl^x ^ 



value of the ordinate at D^ reduces to -^. Further, it is seen 

^ It 



. This stress-moment is negative (according to the conven- 
tion adopted in Art. 146) ; but since /^ is negative, the expres- 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 127 

PI X 
sion — il- is correct in magnitude and sign. For a load P on 

A-j^Ci, distant x from A■^^, the stress-moment is positive, and is 
given in magnitude and sign by the same expression as in the 
preceding cases, — ^. The ordinates at C^ and D^ also have 
the general values 

^^^^~ — u — ' ^^~ /^ ' 

the latter expression being negative (because /^ is negative and 
4~^2 positive) while the former is positive (because 4 and /j — «i 
are both positive). Further, if A^C^ and D^B^ ^^ prolonged, 
they have a common ordinate -^ at the point F^. This ordi- 
nate is negative, agreeing in sign with Z^. 

General result. — It is thus seen that the construction of the 
Influence line may follow one general rule in all cases. This 
rule may be stated as follows, designating points by letters, as 
above : 

From the point F^ erect an ordinate of magnitude and sign 

^-^. From the extremity of this ordinate draw straight lines to 

A^ and B<^, the former intersecting the vertical through C^ in Q, 
the latter intersecting the vertical through D^ in D^. The line 
A^C^D^B^ is the required influence line.* 

148. Approximate Position of Load-Series causing Greatest 
■Stress in a Given Member. — Inspection of the influence dia- 
gram for a given member enables the position of loads causing 
maximum stress to be estimated approximately. 

Member of chord carrying moving loads. — For the member 
C-^D-^ (Fig. 48, PL VIII), the influence diagram {A) shows that 

* The same general rule applies when Fi falls between A^ and Q, or between D-^ and 
B^, Hi being negative in one case, and 11.2 in the other, while both 4 and /o are positive in 
both cases. These cases occur in no ordinary form of truss, except that with subordinate 
bracing. This form is discussed in Art. 160, and the influence diagrams for negative val- 
ues of «i and of «2 are shown on PI. VIII. 



128 GRAPHIC STATICS. 

all loads on the truss produce the same kind of stress, and that 
the effect of a given load is greater the nearer it is * to D-^, 
Therefore, with any given load series, it is to be expected that 
the greatest stress will occur when (a) the truss is heavily 
loaded throughout the span, and (Ji) the heaviest loads are near 
the point D^. 

CJiord not carrying moving loads. — In a similar manner, the 
influence diagram (B) shows that the greatest stress in C{D^ is 
to be expected when {a) the entire span is heavily loaded, and 
{b) the heaviest loads are near D^. 

Web member. — Diagram (C), Fig. 48, PI. VIII, shows that 
for loads between A^ and E^ the stress-moment is positive, 
while for loads between E^ and B^ it is negative. Hence for 
the maximum stress of the kind corresponding to a positive 
stress-moment {a) the portion A^E^ should be heavily loaded 
while little or no load is between E^ and B^, and {h) heavy loads 
should be near 6\. For maximum stress of the kind corre- 
sponding to a negative stress-moment, ia) E-J^-^ should be 
heavily loaded while little or no load is on A-^Ey and (b) heavy 
loads should be f near E)^. 

The above principles serve as an approximate general guide, 
but do not enable us to determine the position of loads for 
maximum stress exactly. A more definite discussion of the 
problem will now be given. 

149. Criterion for Position of Loads causing Maximum or 
Minimum Stress in Any Member of a Truss. — Let a funicular 
polygon and a load polygon be drawn for the given series of 
loads (PI. V). 

As shown in Art. 130, if y denotes any ordinate of the 
funicular polygon, and j/' the corresponding ordinate of the 

* It may, of course, happen that CoQ is greater than DoD^. In such case, the effect 
of a load is greater the nearer it is to Q. 

t The nature of the load-series may be such that a compromise must be made between 
requirements (a) and {b). 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 



129 



load polygon (the axes being LX, L V, and L'X', L' V, 
respectively), 

^'=^1 (■> 

Consider the stress in any member of the truss shown at (A), 

PI. V, as C^D^. Taking a section through C^D^, C^F^, and H-^^F<^, 

the origin of moments for computing the required stress (the 

moment-center) may be taken at F^. Let vertical lines through 

A-^, By C-y D-^ intersect the funicular polygon in A, B, C, D, 

and the load polygon in A\ B' , C , D' . Through F^ draw a 

vertical line intersecting AB in G and CD in F; then (Art. 145) 

the moment of the required stress with respect to the origin F^. 

is 

M=HxFG, 

where H is the pole distance used in drawing the funicular 
polygon. If the load series moves (or if the truss is considered 
as moving while the loads remain stationary), the stress in C-^D-^^ 
is always proportional to FG, and when the stress is a maxi- 
mum or a minimum, the intercept FG is also a maximum or a 
minimum. 

Let the vertical through F-^ intersect A' B' in G' and C'D' in. 
F' . It will now be shown that, as the truss moves horizontally,, 
the change of FG (and therefore of M) per unit distance moved 
is always proportional to F' G' . The reasoning is similar tO' 
that employed in Art. 131. 

Let a, b, c, d, f, g, denote the ordinate s of A, B, C, D, F, G,. 
respectively, measured from the horizontal axis LX ; and a' , b' ,. 
c\ d' , f , g' , the ordinates of A', B' , C', D', F', G', measured 
from the axis L'X'. Also, let 2 denote the abscissa of the point 
y^i, measured from the axis L' Y', and let /]_, /g, /, n-^, n^, n have 
the meanings assigned in Art. 146. 

It may be shown as in Art. 131 that 

cr=-^a + ^-^b- 



130 



GRAPHIC STATICS. 



and by exactly similar reasoning, 

f= — c-] — ^d. 

n n 

Therefore 

M=HxFG=^H{g-f)=H{^ja+-jb--^^c--^^d\. . (2) 

Similar reasoning applied to the lines A'B', CD' leads to the 
equation 

F'G'=g'-f'J-^a' + -^,b'-^^c'--^d' (3) 

The rate of change of J/ as the truss moves horizontally is found 
by differentiating (2) with respect to z. Observing (as in Art. 131) 

that — , -— , ... are the values of -^ 2X A, B, . . ,, equation (i) 
dz dz dx 

shows that 

dz dz 

the differentiation of equation (2) therefore gives 



dz L L 11 n 



Comparing with (3), 



^=g'-f = F'G' (4) 

dz 



It follows from equation (4) that when the position of the 
loads is such that J/ is a maximum or a minimum (that is, when 
the stress in C-^^D-^ is a maximum or a minimum), the ordinate 
F' G' is equal to zero. 

The above reasoning is general,* and the conclusion applies 
to any member of the truss. In the case of chord members the 
point F-^ will in general fall within the span ; only in exceptional 

* That the above reasoning is rigorous for the case of concentrated loads as actually 
applied to the truss in practice follows from the considerations mentioned in the note in 
Art. 132. With slight change in the form of statement, the conclusion holds even in the 
ideal case of concentrated loads in the strict mathematical sense. 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 



131 



forms of truss will it fall without the panel C-^D-^, though it may 
coincide with C^ or D-^. The figure for the case of a web mem- 
ber is shown at {B), PL V, the letters A, B, C, D, F, G, with or 
without subscripts or accents, denoting corresponding points in 
the constructions shown at {A) and at {B). The point F-^ gener- 
ally falls without the span in case of a web member; it is possi- 
ble that it should fall between A^ and C-^ or between D-^ and B^. 
In any case equation (4) holds, and the condition for maximum 
or minimum stress takes the above general form. 

The criterion above reached {F' G' = o) is easy of application. 
Let a strip of cardboard be prepared, showing the truss-skele- 
ton and the moment-centers for all members of the truss. By 
applying this to the diagram showing the load polygon, the 
points A' and B' may be located approximately by inspection 
(an accurate determination of them not usually being necessary 
for the purpose in hand); by stretching a thread through A' 
and B' , the point G' may be located approximately ; then, hold- 
ing one point of the thread at G' , it may be stretched so as to 
pass through one of the points C and D' , and if it passes 
through the other also, the condition for maximum or minimum 
stress is satisfied. As will be shown presently, when the load- 
series consists of concentrated loads, a load will generally be at 
C^ or D^ when the stress is a maximum ; the load polygon will 
therefore be vertical (or very steep) at C or at D' , so that one 
of these points may be shifted vertically without changing 
appreciably the position of the other or of either A' or B'. In 
some cases (but not usually) it may be necessary to locate A', 
B', and G' accurately. 

150. Positions giving Maximum and Minimum Values of the 
Stress Distinguished. — If ^— / and g' —/' have like signs, the 
magnitude of g"—/ increases with a positive displacement, and 
decreases with a negative displacement ; if they have unlike 
signs, £"—/ decreases with a positive displacement, and increases 
with a negative displacement. It follows that, if the truss is in 



132 



GRAPHIC STATICS. 



a position of maximum stress for the member considered, a 
small positive displacement makes the signs of g—f and g' —f 
unlike, while a small negative displacement makes their signs 
like ; while the reverse is true for small displacements from a 
position of minimum stress. (The words " maximum " and 
" minimum " here refer to magnitude, irrespective of the nature 
of the stress.) 

A complete discussion of this principle would require the 
consideration of various forms of truss. It will be sufficient 
to illustrate the application to the form shown on Plates V 
and VI. 

151. Application of General Principles. — On PI. VI is repre- 
sented a truss of 60 ft. span, divided into six equal panels. 
The greatest depth of the truss is 12 ft. ; the lower chord 
(carrying the moving loads at the joints) is straight, while the 
upper joints lie upon the arc of a circle. The two web mem- 
bers intersecting at any upper joint are of equal length and 
slope. 

The moving load consists of the two locomotives and train 
represented on PI. Ill, for which a load polygon and a funicular 
polygon are drawn on PL VI. The diagram represents half the 
actual weight of the train, it being assumed that the load is 
divided equally between two trusses. 

The construction for determining the greatest stress in the 
chord member C-J)^ is shown at {A), PL VI, and the construc- 
tion for the case of the web member HD^ is shown at (^B). Like 
lettering is used in the two cases. 

Mevibev of lozver chord. — From the approximate general rule 
deduced in Art 148, it follows that, in order that the stress in 
C-JD-^ may have its greatest value, the truss must be loaded as 
completely as possible. Moreover, since the effect of a given 
load is greater the nearer it is to the panel C^D-^, the positions 
first to be tested are those bringing the heaviest loads upon or 
near the panel. 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 



133 



The condition for maximum or minimum stress is found to 
be satisfied in the position in which the truss is shown in the 
figure, the heavy load 5 '6' being at D^ ; but not in any other posi- 
tion in which the heaviest loads are upon or near the panel C^D-^ 
and the truss fully loaded. The criterion of Art. 150 shows 
conclusively that this position gives a maximum and not a mini- 
mum value of the stress ; for g—f is positive, and a positive 
displacement of the truss makes g' —f negative, while a nega- 
tive displacement makes it positive.* 

The value of the maximum stress in C-J)^ may now be deter- 
mined. The pole distance is looooo lbs.; FG — g.g ft.; -F^-F^ 
(the moment-arm of the stress) is 12 ft. The required stress is 
therefore a tension of 

g.Q X lOOOOO o^^^^ ^u 

^-^ — = 82500 lbs. 

12 

Web member. — Referring to {B\ PL VI, consider the mem- 
ber HD-^. From Art. 148 it is known that when this member 
sustains its greatest tension the portion D-^B-^ must be as com- 
pletely loaded as possible (with A^C-^ probably free from loads), 
while to produce the greatest compression A-^C-^ must be as 
fully loaded as possible (with D-J3-^ probably free from loads). 
Whether, in either case, loads should be between C-^ and D-^ is 
not known without applying an exact test. 

Considering the case of tension, let the load series be brought 
from the rjght until the foremost load reaches the panel, and 
let the criterion for maximum stress be applied to successive 
positions until one is found in which it is satisfied. It is found 
that the criterion is satisfied when the second load (2-3) is at 
D^. That this position gives a maximum and not a minimum 
value of the stress is found by applying the test of Art. 150. 
For g—f is negative ; a positive displacement of the truss 

* It will be noticed that, with a series of concentrated loads, F'G' will, as a rule, pass 
through zero only when a load passes one of the four points A-^, B-^, Q, D-^. Further, 
when F' is between C and D' , a load at A-^ or B-^ (when F'G' =6) will correspond to a 
minimum stress, while a load at Q or Di will correspond to a maximum. 



134 GRAPHIC STATICS. 

makes g^-f positive, while a negative displacement makes it 
negative.* 

The value of the maximum tension in the member HD^ may 
now be found by the method of moments. The intercept 
i^6' = 8.8 ft; the pole distance is looooo lbs.; the arm F^J 
= 39.6 ft. The required tension is, therefore,! 

8.8 X looooo ^^ „ 

= 22200 lbs. 

39-6 

152. Algebraic Statement of Condition for Maximum or Mini- 
mum Stress. — The general condition above deduced admits of 
simple algebraic expression. For this purpose let- 

Pi = total load on A-^C-^, 
7^2 = total load on D-^B-^, 
Q = total load on CiD^, 
I^= total load on A-^B^=P^ + P^-{-Q. 

Let /j, l^, n-^, «2> ^) ^^ have the meanings assigned in Art. 146, 
their algebraic signs being carefully observed. 

Referring to Fig. (A), PL V, the point G' divides A'B' into 
segments proportional to /j and /g, and F' divides C'F^' into seg- 
ments proportional to ;/j and n^. Also, with the notation of 
Art. 149, it is seen that 

d<-c'=Q; b'-a'=W=P^ + P^+Q. 
Since A'G'B' is a straight line, 

g'-a' _ d'-g' _ d'-a' _ W , 

/i ~ 4 ~'~T~~~r' 

* Here again it is seen that the condition for maximum or minimum stress will, in 
general, be satisfied only when a load is at one of the four points A-^, B-y, Cy, D-^. In this 
case a maximu^n stress of the kind now under consideration will occur only when a load 
is at D-y. 

t These results, being obtained from small-scaled drawings, are not given as accurate 
values of the stresses, the present object being merely to illustrate the method of pro- 
cedure. By careful work, and with drawings made to a suitable scale, a sufficient degree 
of accuracy may be obtained by the above method. 



TRUSS SUSTAINING ANY SERIES OF MOVING LOADS. 



135 



hence £■' —a'=^^, (i) 

^ -£^'=-f (2) 

Similarly, since C'F'D' is a straight line, 

f-c' d' -/' d'-c' Q, 



hence f — c' = ^-^; (3) 

d<-f=^ (4) 

Subtracting equation (3) from equation (i), and (4) from (2), 
and writing P^ and P^ instead of d —a' and b' — d', there result 
the equations * 

^'-/'=^-^-^.^ (5) 

/'-^'=^-l?-^^ (6) 

If the points F' and G' coincide, as is the case when the con- 
dition for maximum or minimum stress is satisfied, g' —f = 0,' 
and equations (5) and (6) may be written f 

Pi+~Q P.A-Q UT 

Although the foregoing discussion has referred directly to the 

* Equations (5) and (6) are not independent ; either may be derived from the other by 
means of the relations P-^-^- P'2.-\- Q = W, 1-^-^-1^ = 1, n^ + 1^2 = '2- 

t It is interesting to compare this result with the condition for maximum or minimum 
bending moment for a beam, given in Art. 132. If the portion of the load polygon from 
C to D' were replaced by the straight line CD' (i.e. if the load Q were uniformly dis- 
tributed over the panel) , the total load from A-^ to F^ would be P^ + — Q, and the total load 
from Fi to B-^ would be 7^2 + — Q- The condition for greatest stress in CiD^ is therefore 
identical with the condition for maximum bending moment at F^^, on the assumption that 
the load Q is uniformly distributed. This interpretation of equation (7) is of use only 
when Fi falls between Ci and Bi, although it holds in a mathematical sense in all cases. 



136 GRAPHIC STATICS. 

case in which F-^ falls between C-^ and D-^, so that l-^, l^, n^, n^ are 
all positive, the result holds also when any of these quantities 
are negative, signs being duly observed in substituting their 
values in equation (7). 

If the member considered belongs to that chord to which the 
moving loads are applied (as at {A), PL V), /^ and /„ will be 
positive, and one or both the quantities it-^ and n^ will be posi- 
tive. If one of the members C-^F^, -^2^1 were vertical, either 
11-^ or //g would be zero. In no truss of ordinary design, without 
subordinate bracing, would n-^ or n^ be negative in case of a 
chord member. 

In case of a member of the chord carrying no moving loads, 
the point F-^ falls at one end of the panel C-J)-^, and either n^ or 
11^ becomes zero. This may be seen by referring to the member 
H^F^, PI. V, (^). 

In case of a web member (as C\//, PI. V, {B)), F-^ falls without 
the panel CiD-^ (making n-^ or n^ negative) ; in the form of truss 
here figured F-^ also falls without * the span A-^B-,^ (making /j or ^ 
negative). In this case equation (7) may be put into more con- 
venient form. Consider, for example, the maximum tension in 
the member D^H, in the truss shown at {B), PI. VI. From the 
general principles stated in Art. 148 it follows that loads on 
F>-^B-^ cause tension in this member, while loads on A-^C-^^ cause 
compression. If possible, therefore, equation (7) should be 
satisfied by a position making P^ zero or small. The equation 
may be written 



Q=" 

11 



fj^~P^-' •.(«) 



and this is to be satisfied by a position making P-^ = o or as 
small as possible. 

Example. — Test whether the general equation (7) or (8) is 
satisfied in the positions shown on PI. VI. 

* A design is possible which will cause F^, for certain web members, to fall within the 
■span but without the panel Q/J^. This does not occur in any ordinary form of truss. 



TRUSS WITH SUBORDINATE BRACING. 



137 



153. Truss with Parallel Chords. — The application of the 
results of the foregoing discussion to the case of trusses with 
parallel chords presents no difficulty.* The only matter calling 
for special remark relates to the case of web members. 

Referring to {B), PL VI, it will be seen that if the chords 
were parallel the point F-^ (and therefore also F and F') would 
fall at infinity. In order that CD' and A' B' might intersect in 
F', these two lines would therefore need to be parallel. The 
condition for maximum stress in a web member therefore 
reduces, in case of parallel chords, to the condition that A' B' 
and CD' shall be parallel. 

Further, the intercept FG and the moment-arm F^J become 
infinite if the chords are parallel, so that the moment-equation 
takes an indeterminate form. The stress in the member HD-^ 
can, however, be easily determined by resolving vertically all 
forces acting upon the portion of the truss to the left of a 
section cutting the member in question and two chord members. 
See Art. 142. 

The algebraic formula expressing the condition for maximum 
stress in a web member (equation (8), Art. 152) reduces to 
simpler form in case of parallel chords. If F-^ passes to infinity, 
L and n-, approach a ratio of equality while — approaches zero ; 
so that the equation becomes, in the limit. 

The load P^ does not enter this equation, but is to be taken as 
small as possible in applying it. 

§ 4. Tniss with Subordi7tate Bracing. 

154. Description of Truss. — In Fig. 49, PI. VIII, is repre- 
sented a form of truss designed to offer points of support for 
the floor system intermediate between the main panel joints. 

* For an exhaustive treatment of this subject consult a paper by Professor Henry T. 
Eddy, Trans. Am. Soc. C. E., Vol. XXII (1890). 



138 GRAPHIC STATICS. 

The effect of the subordinate vertical members a and diagonal 
members b requires some explanation. With the arrangement 
shown in Fig. 49, the main diagonals are tension members and 
the main verticals compression members, while the subordinate 
verticals sustain tension and the subordinate diagonals compres- 
sion. In the form shown in Fig. 50 (PL VIII) the subordinate 
diagonals sustain tension. The members represented by the 
broken lines in both Fig. 49 and Fig. 50 do not act when the 
truss sustains dead loads only. Such members will in general 
be needed only in panels near the middle of the span. In 
Fig. 49 the two upper diagonals (and in Fig. 50 the two lower 
diagonals) in each main panel are designed for tension only ; 
the one represented by the broken line comes into action only 
when the loading is such that the other would be thrown into 
compression, 

155. Determinateness of Stresses. — Disregarding counter- 
braces, it is easily seen that the stresses in all members are 
determinate. 

In the case of members such as e, f, or g (Fig. 49), the deter- 
minateness follows from the fact that any such member can be 
made one of three through which a section may be passed com- 
pletely dividing the truss. 

The stress in ^ is determinate, being always equal to that in g. 
For a like reason, the stress in a is determinate, being always 
equal to the load carried at C\. 

Considering the point of intersection of a, b, d, e, since the 
stresses in a and e are determinate, it follows that the stresses 
in b and d are also determinate ; the force polygon for any four 
forces in equilibrium being completely determinate when two of 
the forces are known completely, and the directions of the other 
two are given. This shows that the stresses in b and d would 
be determinate even if d and e were not collinear. 

Like reasoning applied to an upper joint shows that the 
stress in any main vertical h is determinate. 



TRUSS WITH SUBORDINATE BRACING. 139 

156. Effect of Subordinate Braces on Stresses in Main Mem- 
bers. — Upper cho7'd. — It may be shown that the stresses in the 
upper chord members have, for any position of the loads, the 
same values they would have if the subordinate members were 
not present. Thus, applying the method of moments in the 
usual manner, the center of moments for the member /* is 
the point D-^. The stress is determined by dividing the truss 
by a section through/, e, and ^, and applying the conditions of 
equilibrium to either portion of the truss. For a load between 
A-^ and F^, consider the right portion of the truss. The reaction 
at B-^ is the same, whatever the division into panels ; hence the 
stress is the same as if the members a and b were not present. 
For a load between B-^ and D-^, the same conclusion is reached 
by considering the left portion of the truss. For a load between 
F-^ and D-^ the moment equation for the right portion of the 
truss will contain the reaction at B-^, and the portion of the load 
carried at D-^ ; and this portion is changed by the subdivision 
of the panel F^D-^. The moment of the load at D-^ is, however, 
zero ; so that the subdivision of the panel has no effect upon 
the moment equation nor upon the stress in the member/". 

Web 7neinbers. — At any upper joint of the truss four forces are 
in equilibrium, — two due to the chord stresses and two to the 
stresses in the web members. Any two of these being given, 
the other two are determined. Since the chord stresses have 
the same value whether the subordinate members are present 
or not, the same is true of the web stresses. In determining 
the stresses in d and //, the subordinate members a and b may 
therefore be disregarded. 

The stress in e is changed by the presence of the subordinate 
members if any load is between F-^ and D-^. The value of the 
stress, may, however, be determined by the method of sections, 
the center of moments being at the intersection of / and g. 
The construction shown at {B), PI. VI, may be applied to the 

* The present discussion refers always to Fig. 49, PI. VIII. 



140 



GRAPHIC STATICS. 



member e, the short panel C-J)^ (Fig. 49) corresponding to the 
panel C-JD-^ in PL VI, and the point F-^ being at the intersection 
of/ and ^. 

Lozver chord. — The analysis just applied to the member e is 
applicable also to a lower chord member, such as g. Taking a 
section through/", e, and^, the origin of moments is to be taken 
at the intersection of e and / The construction shown on 
PI. VI may therefore be applied, the panel marked C-JD^ in 
Fig. 49 corresponding to C-J)^ in PI. VI. 

Since the load at C-^ and the member a are collinear, the 
stress in g^ is always equal to that in g. 

157. Condition for Maximum Stress in Main Truss Members. — 

From the results of Art. 156, it follows that the method already 
explained for determining what position of the loads causes a 
maximum stress may be applied to all the main members of the 
truss shown in Fig. 49. The condition in all cases is that the 
points F' and Q , determined as on PI. V or PL VI, shall coin- 
cide. In case of certain members, however, the position of the 
line CD' is changed by the presence of the subordinate mem- 
bers. To summarize : 

In the case of the members / d, and h (Fig. 49), the subordi- 
nate members are to be disregarded in applying the condition 
for maximum stress ; the whole panel F^D-^ takes the place of 
C-^D^ in the construction on PL VI. This principle applies to 
all upper chord members, all main verticals, and the upper por- 
tions of all main diagonals. 

In the case of the members e and g, the short panel marked 
C-J)-^ in Fig. 49 takes the place of C-JD^ in the construction on 
PL VI ; this construction being otherwise unchanged. This 
applies to all lower chord members, and to the lower portions 
of the main diagonals. 

There is no difificulty in applying similar reasoning to the 
truss shown in Fig. 50. 



TRUSS WITH SUBORDINATE BRACING. 



141 



158. Subordinate Members. — When the position of the load 
series is known, the stresses in the subordinate members {a and 
b, Fig. 49) are easily determined. 

The stress in a is always equal to the load carried at C\. 

The members d and e being collinear, the resolved parts 
of the stresses in a and b perpendicular to d and e must be 
equal ; the stress in b can therefore be found directly from that 
in a. 

It remains to determine the position of the load series which 
makes the load at C^ (and therefore the stresses in a and b) a 
maximum. 

Assuming that a load on any panel is carried to the truss 
wholly at the ends of that panel, the condition for maximum 
load at any joint may be deduced without difBculty. 

In Fig. 51 are represented the load polygon and funicular 
polygon for a series of loads, A^F-^ and F-^B-^ being consecutive 




Fig. 51 



panels of any lengths. Vertical lines through A-^, B^, and F^^ 
intersect the funicular polygon in A, B, F, and the load polygon 
in A' , B', F'. The portion of the loads on A^F^ which is car- 



142 



GRAPHIC STATICS. 



ried at F-^ is found by drawing the closing string AF, and 
parallel to it the ray OK' ; the portion carried at F-^ is K'2. 
Similarly, drawing OK" parallel to the closing string FB, 2 K" 
represents the portion of the loads on F-^B^ which is carried at 
F^. The total load carried at F^ is therefore K'K" . 

Prolonging BF to L, AFL and K'OK" are similar triangles; 
hence AL and K' K" vary in the same ratio, as the load series 
moves. But if G is the point in which a vertical through F 
intersects AB, it is seen that FG varies as AL. Therefore, 
when K'K" is a maximum, FG is a maximum. 

Reasoning as in Arts. 131 and 132, it is found that when FG 
is a maximum or a minimum the points A', F' , and B' are 
collinear. 

The condition for greatest load at F-^ is therefore exactly 
analogous to the condition for greatest bending moment at F^ 
on the supposition that A-^B-^ is a simple beam supported at 
A-^ and B^. 

From this analogy the conditions for greatest stresses in the 
members a and b (Fig. 49) may be stated as follows : * 

{a) The panels F-^C-^ and C-^D^ must be loaded as heavily as 
possible, the heaviest loads being near C^. 

{b) The loads on the panels must be proportional to their 
lengths. 

In satisfying requirement {b), a load must generally be at 6\, 
and must be regarded as divided between the panels in some 
arbitrary manner. 

159. Effect of Counterbraces. — The foregoing discussion and 
conclusions need modification when the loading is such as to 
bring counterbraces into action. It is needful to determine 
what counters are needed, what maximum stresses they sustain, 
and whether the maximum stresses in any other members are 
changed by their action. Let it be assumed that when the 

*See paper by Professor H. T. Eddy, Trans. Am. Soc. C. E., Vol. XXII. 



TRUSS WITH SUBORDINATE BRACING. 



143 



train moves from right to left the members called into action 
are those represented in Fig. 52. The conclusions drawn in 
Art. 155 as to the determinateness of the stresses are equally 
applicable to the present case ; but the fact that the mem- 
bers d and e are not collinear makes it necessary to modify 




Fig. 52 

the methods of determining greatest stresses in certain 
members. 

In case of the members /, g^ e, and a (Fig. 52) the usual 
methods obviously apply. 

If b is designed for tension only, it is necessary to determine 
whether it acts when the stress in d is greatest. If it does not, 
the discussion of d is obvious. If b does act, the stress in d is 
not the same as in the case in which d and e are collinear. It 
may be shown, however, that the stress in d always bears a 
constant ratio to that which would exist if d and e were collinear. 
This becomes evident if the force polygon be drawn for the 
forces due to the stresses in /, /', d, and h, and the effect of 
changing the direction of d be observed ; it being remembered 
that such change does not affect the stresses inf,/'. 

In a truss of long span, the stress in d will usually be small, 
since the dead load will be so great as to nearly or quite coun- 
terbalance the greatest live load effect. Counters will be needed, 
if at all, only in panels near the middle of the span. 

The determination of the maximum stress in b with the 
counter d in action is less simple than it would be if d and 
£ were collinear; but the member b will have its true maxi- 
mum when the member d does not act, the train approaching 
from the left. 



144 



GRAPHIC STATICS. 



1 60. Influence Lines. — From the foregoing discussion it is 
seen that the influence lines for the main truss members may be 
drawn by the general method given in Art. 147. In one case, 
however, there is a peculiarity which is worthy of special notice. 

Considering the lower chord member C-JD-^ (Fig. 49, PL VIII), 
the point F-^ falls between A-^ and C-^, making n-^ negative. The 

ordinate F^f^ is made equal to -^, positive ; ^2/2 ^^^ ^-^2 ^^^ 

drawn, intersecting verticals through C^ and D^ in Q and Z^g, 
respectively ; the influence line is A^C^D^B^. This peculiar 
form does not occur in any common form of truss without 
subordinate bracing. 

Figure 50 shows the influence line for the upper chord mem- 
ber^; the subordinate braces being so placed that the diago- 
nals are in tension. Here n^ is negative, F^ falling between 
Z?i and B-^. 

For the subordinate members the influence lines may be 
drawn without difficulty from the principles of Art. 158. 

161. Application. — On PI. VII is shown a truss of 300 fL 
span, divided into ten panels of 30 ft. each. The lengths of the 
main verticals are 30 ft., 45 ft., and 60 ft. The subordinate 
diagonals are so placed as to sustain tension. Counterbraces 
are shown in the central panel only ; a full computation of 
stresses due to live and dead loads will determine whether other 
counters are needed. The moving load is the same as shown 
on PI. III. 

At {A), PI. VII, is shown the position of the truss causing 
greatest stress in the upper chord member C-^F^ , and at {B) the 
position causing greatest stress in the web member C^D{. The 
letters A, B, C, D, F have the same meanings as on Plates V 
and VI, and the construction is self-explanatory. If the lower 
part of the web member C-^K, in Fig. {B), were under consid- 
eration, the point F-^ would be as shown in the figure, but C^D-^ 
would be the double panel C^K (the same as if the subordinate 



UNIFORMLY DISTRIBUTED MOVING LOAD. 



145 



members were omitted). These statements are in accordance 
with the foregoing discussion. 

In each of the cases shown on PI. VII it will be seen that the 
lines A'B' and CD' intersect at a point F' lying in the vertical 
through F^. 

The stresses corresponding to the positions thus determined 
may be computed by the method of moments, as on PI. VI. 
The construction is not shown ; but satisfactory results may be 
secured by careful work and by the use of suitable linear and 
force scales. 

The construction for determining the position for greatest 
stress in a subordinate member is not shown ; it is identical with 
the construction shown on PL IV for determining the position 
for greatest bending moment at a given section of a beam. 

§ 5. Uniformly Distributed Moving Load. 

162. Application of General Method. — The case in which the 
moving load is assumed to have a uniform horizontal distribu- 
tion involves no principles additional to those already explained. 
The foregoing discussion applies to any load distribution what- 
ever. It is, however, of interest to notice the special form 
assumed by the general principles when the load is uniformly 
distributed. 

The load polygon becomes a straight line whose slope depends 
upon the intensity of the loading (the load per unit length). 
The funicular polygon becomes a continuous curve, being in 
fact a parabola with axis vertical, the position of the vertex 
depending upon the position of the pole of the force diagram. 
For the purpose of drawing the funicular polygon the load may 
be replaced by a series of concentrated loads ; the sides of the 
resulting polygon will be tangent to the desired curve, and the 
polygon will coincide more and. more nearly with the desired 
parabola, the smaller and nearer together the concentrated loads 
are taken. 



146 GRAPHIC STATICS. 

It is obvious that for maximum chord stresses the only condi- 
tion is that the moving load shall cover the whole span. 

For maximum stress in any web member, the head of the 
load series must be at some point in the corresponding panel ; 
the exact position may be determined by applying the general 
method. The position of the head of the load series is shown 
at a glance by the influence diagram. Thus, from diagram {C), 
Fig. 48 (PI. VIII), it is obvious that the greatest tension in the 
member C^'D-^^ will occur when the load covers EiB^ completely, 
£iA-^ being free from load. 

163. Assumption of Full Panel Loads. — The determination 
■of the maximum web stresses is somewhat simplified by an 
approximate assumption as to the way in which the uniformly 
distributed load comes upon the truss. Thus, in Fig. 53, let it 
be required to determine the greatest tension in the member 
BB' . The moving load must be brought on from the right until 
its head is at some point between A and B. In this position, 
the loads actually supported at the various joints are as follows: 

A' 




At C a full panel load (half the load between B and C, and half 
that between C and D) ; at each of the points D, E, and F also 
a full panel load ; at B less than a full panel load (half the load 
•on BC, and a portion of that on AB). Let it be assumed that, 
when the tension in BB' is greatest, all joints from B to /^inclu- 
sive sustain full panel loads, while all joints to the left of B 
(only one in the case shown) are free from loads. 

If a similar assumption is made in case of every chord mem- 
ber, the resulting stresses will not differ greatly from those 
obtained by rigorously adhering to the assumption of a uniformly 
distributed load, the error being on the side of safety. It is to 



UNIFORMLY DISTRIBUTED MOVING LOAD. 



147 



be remarked, also, that the case of a strictly uniform dis- 
tribution is not realized in practice, so that the results of 
the above assumption are probably as reliable as would be 
obtained by following out strictly the assumption of uniform 
•distribution. 

The method of determining maximum stresses, on the above 
assumption as to the loading, will be illustrated by reference to 
the truss shown in Fig. 54, PI. VIII. In this truss the panels 
are of equal length ; but the method applies also to the case of 
unequal panels, the only difference being that the " full panel 
loads " for different joints are not equal unless the panels are 
equal. 

The principle of counterbracing will be employed here, the 
diagonals being constructed to sustain tension only. 

164. Dead Load Stresses. — The dead load stresses may 
be determined by means of a stress diagram, as in the roof- 
truss problems already treated. Two points should be ob- 
served in drawing this diagram, (i) If dead loads are taken 
to act at upper as well as at lower joints, the force poly- 
gon for the loads and reactions must show these forces in 
the same order as that in which their points of application 
occur in the perimeter of the truss. (2) The diagonal mem- 
bers assumed to act are taken as all sloping in the same 
•direction. 

The reason for the first point is the same as already explained 
in Art. 90; viz. that unless the forces be taken in the order 
mentioned, the stress diagram cannot be the true reciprocal of 
the truss diagram and certain lines will have to be duplicated. 
The reason for assuming the diagonals as all sloping in the 
same way is the same as in the case of the roof truss with 
counterbracing (Arts, iii and 112). 

The dead-load stress diagrarn is not shown, since its con- 
struction involves no principle not already fully explained and 
illustrated. 



148 GRAPHIC STATICS. 

165. Stresses Due to Moving Load. — {a) Chord members. — • 
By applying the method of sections it is easily seen that a load 
at any point produces tension in every lower chord member and 
compression in every upper chord member. It follows that the 
greatest stresses in the chord members will occur when the truss 
is fully loaded. A convenient method of determining the live 
load stresses is, therefore, to draw a stress diagram, as in case of 
fixed loads. This diagram is not shown. 

ih) Web members. — In case of a web member, loads in dif- 
ferent positions tend to cause opposite kinds of stress. Thus, 
considering the member /';/ (Fig. 54, PI. VIII), a tension is 
caused in it by a load at either of the points ab, be, cd, de, or ef, 
while a compression is caused by a load at fg, g/i, or hi. Simi- 
larly, loads at ab, be, cd, de, and ef all tend to throw compression 
on the vertical member /';«', while loads on fg, gh, hi have the 
opposite tendency. 

Therefore, to produce the greatest tension upon /'«' (and 
compression on fin'), the live load must act only at ef and all 
joints to the right; while to cause the greatest compression on 
f'n' (and tension upon f'm') the live load must act only at fg, 
gh, and hi. A similar statement will hold regarding any other 
web member. Since counterbraces are to be used in all panels 
in which diagonal members would otherwise be thrown into 
compression, we shall need only to consider the greatest tension 
in each diagonal and the greatest compression in each vertical. 
We shall first outline the method to be employed, and then 
explain the construction. 

To determine the greatest tension in a diagonal member, as 
g'm' : Assume the live load to come upon the bridge from the 
right until there are full loads at the joints ab, be, cd, de, ef, and 
fg. Take a section cutting g'm' and the two chord members 
gg' and m'm, and consider the forces acting upon the portion of 
the truss to the left of the section. These forces are four in 
number: the reaction at the support and the forces acting in the 
three members cut. Hence we first determine the reaction. 



UNIFORMLY DISTRIBUTED MOVING LOAD. 



149 



and then determine the three other forces for equiHbrium by 
the method of Art. 42. 

The construction is shown in Fig. 54 (PI. VIII). ABCDEFGHI 
is the force polygon for the eight live loads that may come 
upon the truss. Choosing a pole O, the funicular polygon 
for the eight loads is next drawn. Now, turning the attention 
to the member ^'m', the loads ^/^ and /«'are assumed not to act. 
The reactions at the supports for this case of loading are found 
in the usual way. Prolong oa and og to intersect the lines 
of action of the two reactions, and join the two points thus 
determined. This gives the closing line of the funicular poly- 
gon (or oni). The ray OM is now drawn parallel to the string 
cm, and the two reactions are (7 J/ and MA. 

Now take a section through mm' , m'g' , and g'g, and apply 
the construction of Art. 42 to the determination of the forces 
acting in the three members cut. The resultant of GM and 
the force in gg' must act through the point X (the intersection 
of gg' produced and ij). The resultant of the forces in vim' 
and mg' must act through their intersection Y. Hence these 
two resultants (being in equilibrium with each other) must both 
act in the line XY. From G draw a line parallel to gg' , and 
from M 2i line parallel to XY; mark their point of intersection 
G' . Then MG' is the resultant of the forces in mvi' and 7)i'g' . 
From J/ draw a line parallel to mm' , and from G' a line parallel 
to the member g'm' , and mark their point of intersection M' . 
Then M'G' represents the force in the member m'g'. This is 
also the value of the greatest stress in m'g'. 

To determine the stress in the vertical member /'g', the 
same loading must be assumed, and a similar construction is 
employed. Take a section cutting //', /'g', and g'g, and deter- 
mine forces acting in these three lines which shall be in equi- 
librium with the reaction GM. This reaction is in equilibrium 
with MG' and G'G, the former having the line of action XY. 
Then MG' is resolved into two forces having the directions of 
the members g'/' and /'/. The stress in g'/' is found to be 



I50 GRAPHIC STATICS. 

a compression, represented in the stress diagram by the line 
G'L'. 

The maximum live load stress in every web member may be 
found in the same way. If the above reasoning is understood,, 
there will be no difficulty in applying the same method to the 
remaining members. 

1 66. Maximum Stresses. — By combining the stresses due to 
live and dead loads, the maximum stresses are easily deter- 
mined. 

Web members. — When the web members are considered, the 
effect of counterbracing needs careful attention. 

The construction above explained gives the greatest live load 
tension in each diagonal. It may be that for certain members,, 
this tension is wholly counterbalanced by the dead loads. In 
any panel in which this is the case, the member shown will 
never act and may be omitted. The counterbrace must then 
be considered. 

Evidently, the algebraic sum of the stresses due to live and 
dead loads will be the true maximum tension in each of the 
diagonals shown. For the greatest tension in the other system 
of diagonals, the load must be brought on the bridge from the 
left; or, what amounts to the same thing, the tension already 
found for any one of the diagonals shown is also the greatest 
tension in the diagonal sloping the opposite way in the panel 
equally distant from the middle of the truss. (In fact, the 
stresses in all members due to a movement of the loads from 
left to right may be obtained from the results already reached, 
by consideration of symmetry.) 

As to the vertical members, two values of the stress must be 
compared in every case — namely, the greatest compressions 
corresponding to the two directions of the moving load. But 
both can be obtained from the above results by considering the 
symmetry of the truss. For example, the stress found for I'g' 
is a possible value for the stress in r'b' , and must be compared 



UNIFORMLY DISTRIBUTED MOVING LOAD. 



151 



with the value obtained for the latter member when the load 
moves from right to left. It is possible, also, that certain of 
the verticals may be in tension when the dead loads act 
alone. 

Maximum chord stresses. — These are found by combining 
the stresses due to fixed and moving loads, determined as 
already described. 



Part III. 



CENTROIDS AND MOMENTS OF INERTIA. 



CHAPTER VIII. CENTROIDS. 



§ I. Centroid of Parallel Forces. 



167. Composition of Parallel Forces. — The composition of 
complanar parallel forces can always be effected by means 
of the funicular polygon, by the method of Art. 27. It is now 
necessary to consider parallel systems more at length, as a pre- 
liminary to the discussion of graphic methods for determining 
•centers of gravity and moments of inertia. 

168. Resultant of Two Parallel Forces. — Let ab and be (Fig. 
55) be the lines of action of two parallel forces, their magni- 
tudes being AB and BC (not shown). Let 
ac be the line of action of their resultant, 
and AC (not shown) its magnitude. By 
the principle of moments (Art. 50) the sum 
of the moments of AB and BC about any 
point in their plane is equal to the moment 

of ^C about the same point. If the origin of moments is on 
ac, the moment of AC 15 zero; and therefore the moments of 
AB and BC are numerically equal (but of opposite signs). 

Let any line be drawn perpendicular to the given forces, 
intersecting their lines of action in P', Q', and R' respectively. 
Then ABxP'R' = BCx Q'P'. 



ab 



R'. 



IPig. S5 



152 



CENTROID OF PARALLEL FORCES. 



153 



Let any other line be drawn intersecting the three lines of 
action in P, Q, and R respectively. Then 

PR ^ QR 
P'R' Q'R'' 

and therefore ABxPR = BCxQR. 

That is, PQ is divided by the line ac into segments inversely 

proportional to AB and BC. 

If AB and BC act in opposite directions, the line ac will be 
outside the space included between ab and be ; but the above 
result is true for either case. 

169. Centroid of Two Parallel Forces. — If the lines of action 
nb and be (Fig. 55) be turned through any angle about the points 
P and Q respectively, the forces remaining parallel and of 
unchanged magnitudes, the line of action of their resultant will 
always pass through the point R. For, by the preceding article, 
the line of action of the resultant will always intersect PQ in a 
point which divides PQ into segments inversely proportional to 
AB and BC. Hence, if AB and BC remain unchanged, and 
also the points P and Q, the point R must also remain fixed. 

If P and Q are taken as the points of application of AB and 
BC, R may be taken as the point of application of AC, in 
whatever direction the parallel forces are supposed to act. The 
point R is called the centroid * of the parallel forces AB and 
BC for the fixed points of application P and Q. 

170. Centroid of Any Number of Parallel Forces. — Let AB, 
'BC, and CD be three parallel forces, and let P, Q, and S 

(Fig. 56) be their fixed points of 
application. Let R be the cen- 
troid of AB and BC, and AC ihtxr 
resultant. Take R as the fixed 
point of application of AC, and 
determine T, the centroid oi AC 



G a 
Q 



Fii 



je 



* The name center of parallel forces has been quite commonly used instead of centroid 
as above defined. The latter term has, however, been adopted by some of the later writers, 
and seems on the whole a better designation. 



154 GRAPHIC STATICS. 

and CD. Let AD be the resultant of AC a.nd CD; then AD is 
also the resultant of AB, BC, and CD. 

Now if AB, BC, and CD have their direction changed, but 
still remain parallel and unchanged in magnitude, it is evident 
that the point T, determined as above, will remain fixed and will 
always be on ad, the line of action of AD. The point 7" is called 
the centroid of the three forces AB, BC, and CD. 

By an extension of the same method, a centroid may be 
determined for any system of parallel forces having fixed points 
of application. Hence the following definition may be stated : 

The centroid of a system of parallel forces having fixed points 
of application is a point through which the line of action of 
their resultant passes, in whatever direction the forces be taken 
to act. 

In determining the centroid by the method just described, 
the forces may be taken in any order without changing the 
result. For the centroid must lie on the line of action of the 
resultant ; and since this is a determinate line for each direction 
in which the forces may be taken to act, there can be but one 
centroid. 

171. Non-complanar Parallel Forces. — The reasoning of the 
preceding articles is equally true, whether the forces are com- 
planar or not. In what follows we shall deal either with 
complanar forces, or with forces whose points of application are 
complanar. No more general case will be discussed. 

172. Graphic Determination of Centroid of Parallel Forces.^ 

If the line of action of the resultant of any system of parallel 
forces be found for each of two assumed directions of the forces, 
the point of intersection of these two lines is the centroid of 
the system. Moreover, if the points of application are com- 
planar, the two assumed directions may both be such that the 
forces will be complanar. 

Thus, let the plane of the paper be the plane containing the 
given points of application, and let ab, be, cd, de, ef (Fig. 57) be 



CENTROID OF PARALLEL FORCES. 



155 



the points of application of five parallel forces, AB, BC, CD, 
DE, EF. Draw through these points parallel lines in some 
chosen direction, and taking them as the lines of action of the 
given forces, construct the force and funicular polygons corre- 




sponding. The line of action of the resultant is drawn through 
the intersection of the strings oa and of, and this line contains 
the centroid of the given forces. Next draw through the given 
points of application another set of parallel lines, preferably per- 
pendicular to the set first drawn, and draw a funicular polygon 
for the given forces with these lines of action. (It is unneces- 
sary to draw a new force diagram, since the strings in the 
second funicular polygon may be drawn respectively perpendic- 
ular to those of the first.) This construction determines a 
second line as the line of action of the resultant corresponding 
to the second direction of the forces. The required centroid of 
the system is the point of intersection of the two Hues of 
action of the resultant thus determined, and is the point af in 
the figure. 

Example. — Find graphically the centroid of the following 
system of parallel forces, and test the result by algebraic com- 
putation : A force of 20 lbs. applied at a point whose rectangu- 
lar coordinates are (4, 6) ; 12 lbs. at the point (12, 3) ; 20 lbs. at 



156 GRAPHIC STATICS. 

the point (10, 10) ; —10 lbs. at the point (7, —9); —8 lbs. at 
the point (—5, — 10). 

173. Centroid of a Couple. — If AB and BC are the magni- 
tudes of any two parallel forces applied at points P and Q, 
then by Art. 169 their centroid R is on the line PQ and is 
determined by the equation 

PR^BC 
QR AB' 

If AB and BC are equal and opposite forces, PR and QR must 
be numerically equal, and R must be outside the space between 
the lines of action of AB and BC. These conditions can be 
satisfied only by making PR and QR infinite. Hence we may 
say that the centroid of two equal and opposite forces lies on 
the line joining their points of application and is infinitely dis- 
tant from these points. 

174. Centroid of a System whose Resultant is a Couple. — 

If a system with fixed points of application is equivalent to a 
couple, its centroid will be infinitely distant from the given 
points of application. A line containing this centroid can be 
determined as follows : 

Take the given forces in two groups ; the resultants of the 
two groups will be equal and opposite. Find the centroid of 
each group, and suppose each partial resultant applied at the 
corresponding centroid. Then the centroid of the whole sys- 
tem is the same as that of the couple thus formed, and will lie 
on the line joining the two partial centroids. 

If the separation into groups be made in different ways, dif- 
ferent couples and different partial centroids will be found. 
The different couples are, of course, equivalent ; and it may be 
proved that the lines joining the different pairs of partial cen- 
troids are all parallel, and intersect in the (infinitely distant) 
centroid of the whole system. 

For, suppose the given system to be equivalent to a couple 



CENTER OF GRAVITY. 



157 



Q with points of application A and B, and also to a couple Q' 
with points of application A' and B'. These two couples must 
be equivalent to each other, whatever be the direction of the 
forces. Let AB be taken as this direction. The two equal 
and opposite forces of the couple Q neutralize each other, since 
their lines of action are coincident ; hence the two forces of 
the couple Q' must also neutralize each other. Therefore A'B' 
must be parallel to AB. 

175. Moment of a Force about an Axis. — The moment of a 
force with respect to a given axis, as defined in Art. 47, 
depends not only upon the point of application of the force, 
but upon its direction. In dealing with systems of forces 
whose direction may change but whose points of application 
are complanar, we shall need to compute moments only for 
axes lying in the plane of the points of application ; and the 
forces may usually be regarded as acting in lines perpendicular 
to this plane. Hence we shall compute moments by the 
following rule : 

The moment of a force with reference to any axis is the 
product of the magnitude of the force into the distance of its 
point of application from the axis. 

§ 2. Center of Gravity — Definitio7is and General Principles. 

176. Center of Gravity of Any Body. — Every particle of 
a terrestrial body is attracted by the earth with a force propor- 
tional directly to the mass of the particle. The resultant of 
such forces upon all the particles of a body is its weight ; and 
the point of application of this resultant is called the center of 
gravity of the body. The lines of action of these forces may 
be assumed parallel without appreciable error. We may there- 
fore define the center of gravity of a body as follows : 

If forces be supposed to act in the same direction upon all 
particles of a body, each force being proportional to the mass 



158 GRAPHIC STATICS. 

of the particle upon which it acts, the centroid of this system 
of parallel forces is the center of gravity of the body. 

This point is also called center of mass, and center of inertia, 
either of which is a better designation than center of gravity. 
The latter term is, however, in more general use. 

177. Centers of Gravity of Areas and Lines. — The term 
center of gravity, as above defined, has no meaning when 
applied to lines and areas, since these magnitudes have no 
mass, and hence are not acted upon by the force of gravity. 
It is, however, common to use the name center of gravity in 
the case of lines and areas, with meanings which may be stated 
as follows : 

The center of gravity of an area is the center of gravity of 
its mass, on the supposition that each superficial element has a 
mass proportional to its area. This point would be better 
described as the center of area. 

The center of gravity of a line is the center of gravity of its 
mass, on the assumption that each linear element has a mass 
proportional to its length. The term center of lengtJi is pref- 
erable, and will often be used in what follows. 

Similar statements might be made regarding geometrical 
solids, but we shall have to deal chiefly with lines and areas. 

178. Moments of Areas and Lines. — Definition. — The mo- 
ment of a plane area with reference to an axis lying in its plane 
is the product of the area by the distance of its center of 
gravity from the axis. 

Proposition. — The moment of any area about a given axis is 
equal to the sum of the moments of any set of partial areas 
into which it may be divided. For, by the definition of center 
of gravity, a force numerically equal to the total area and 
applied at its center of gravity is the resultant of a system of 
forces numerically equal to the partial areas and applied at 
their respective centers of gravity ; and the moment of any 



CENTER OF GRAVITY. 1 59 

force is equal to the sum of the moments of its components 
{Art 50). 

A similar definition and proposition may be stated regarding 
lines. 

The moment of an area or line is zero for any axis containing 
its center of gravity. 

179. Symmetry. — Two points are symmetrically situated 
with respect to a third point if the line joining them is bisected 
by that point. 

Two points are symmetrically situated with respect to a line 
or plane when the line joining them is perpendicular to the 
given line or plane and bisected by it. 

A body is symmetrical with respect to a point, a line, or a 
plane, if for every point in the body there is another such that 
the two are symmetrically situated with respect to the given 
point, line, or plane. The point, line, or plane is in this case 
called a point of symmetry, an axis of symmetry, or a plane of 
symmetry of the body. 

180. General Principles. — (i) The center of gravity of two 
masses taken together is on the line joining the centers of 
gravity of the separate masses. For it is the point of applica- 
tion of the resultant of two parallel forces applied at those 
points. 

(2) If a body of uniform density has a plane of symmetry, 
the center of gravity lies in this plane. If there is an axis of 
symmetry, the center of gravity lies in this axis. If the body 
is symmetrical with respect to a point, that point is the center 
of gravity. For the elementary portions of the body may be 
taken in pairs such that for each pair the center of gravity is 
in the plane, axis, or point of symmetry. 

181. Centroid. — The center of gravity of any body or geo- 
metrical magnitude is by definition the same as the centroid of 
a certain system of parallel forces. It will be convenient. 



l60 GRAPHIC STATICS. 

therefore, to use the word centroid in most cases instead of 
center of gravity. 

§ 3. Centroids of Lines and of Areas. 

182. General Method of Finding Centroid. — The centroid 
of any area may be found by the following method : Divide the 
given area into parts such that the area and centroid of each 
part are known. Take the centroids of the partial areas as the 
points of application of forces proportional respectively to those 
areas. The centroid of this system of forces is the centroid of 
the total area, and may be found by the method of Art. 172. 

The centroid of a line may be found by a similar method. 

The method just described is exact if the magnitudes of the 
partial areas and their centroids are accurately known. If the 
given area is such that it cannot be divided into known parts, it 
will still be possible to get an approximate result by this 
method. 

In applying this general method, it is frequently necessary to 
know the centroids of certain geometrical lines and figures, and 
also the relative magnitudes of the areas of such figures. 
Methods of determining these will be given in the following 
articles. 

183. Centroids of Lines. — The centroid of a straigJit line is 
at its middle point. 

Broken line. — The centroid of a broken line is the center of 
a system of parallel forces, of magnitudes proportional to the 
lengths of the straight portions of the line, and applied respec- 
tively at their middle points. It may be found graphically by 
the method of Art. 172, or by any other method applicable to 
parallel forces. 

Part of regular polygon. — For the centroid of a part of a 
regular polygon, a special construction is found useful. 



CENTROIDS OF LINES AND OF AREAS. 



i6i 




Let ABCDE (Fig. 58) be part of such a polygon, and O the 
center of the inscribed circle. Let 
r = radius of inscribed circle ; / = 
length of a side of the polygon ; s 
= total length of the broken line 
AE. Through O draw OC, the axis 
of symmetry of AE\ and MN, per- 
pendicular to OC. 
First, the centroid must lie on OC. 

Second, to find its distance from O, assume a system of equal 
and parallel forces applied at the middle points of the sides AB, 
BC, etc. The required centroid is the point of application of 
the resultant of these forces. Taking MN as the axis of 
moments, and letting x = required distance of centroid from 
MN, and x-^, x^, x^, x^ the distances of the middle points of AB,. 
BC, etc., from MN, we have from the principle of moments, 

Ixi + /;r2 + /^3 + IXi^ = sx. 

But if Ab and Bb be drawn perpendicular respectively to PQ 
and MNwQ have from the similar triangles ABb, POQ, 

AB^RO J_^r 
Ab PQ^^ Ab xi 



In the same way, 



.•. Ix^^r ' Ab. 

1x2 = r • be, 

lx^=r- cd, 

lXi = r-dE. 

Hence, s-x=r{Ab-{-bc-\-cd+dE) = r-AE, 

where AE is equal to the projection of the broken line ABCDE 
on MN 

The centroid G may now be found graphically as follows : 
Make Oc' = r\ ON=^s; OE' = ^ AE; draw W^'.' Then G is. 
determined by drawing E' G parallel to Nc' 



l62 GRAPHIC STATICS. 

Circular arc. — The above construction holds, whatever the 
length of the side /. If this length be decreased indefinitely, 
while the number of sides is increased indefinitely, so that the 
length s remains finite, we reach as the limiting case a circular 
arc. The same construction therefore applies to the determina- 
tion of the centroid of such an arc, r denoting the radius of the 
circle and s the length of the arc. 

184. Centroids of Geometrical Areas. — Parallelogram. — The 
centroid of a parallelogram is on a line bisecting two opposite 
sides. 

Let ABCD (Fig. 59) be a parallelogram, and EF a line 
bisecting AD and BC. Divide AB into any even number of 

equal parts, and through the points 
of division draw lines parallel to BC. 
Also divide BCm.X.0 any even number 
of equal parts and draw through the 
points of division lines parallel to 
AB. The given parallelogram is thus 
divided into equal elements. Now 
consider a pair of these elements, such as those marked P and 
Q in the figure, equally distant from AD, and also equally 
distant from EF, but on opposite sides of it. The centroid of 
the two elements taken together is at the middle point of the 
line joining their separate centroids. If the number of divisions 
of AB and of BC be increased without limit, the elements 
approach zero in area, and the centroids of P and Q evidently 
approach two points which are equally distant from EF. Hence 
in the limit, the centroid of such a pair of elements lies on the 
line EF. But the whole area ABCD is made up of such pairs ; 
hence the centroid of the whole area is on the line EF. For 
like reasons it is also on the line bisecting AB, and DC; hence 
it is at the intersection of the two bisectors. 

The point thus determined evidently coincides with the point 
of intersection of the diao:onals AC and BD. 




CENTROIDS OF LINES AND OF AREAS. 



163 




Triangle. — The centroid of a triangle lies on a line drawn 
from any vertex to the middle of the opposite side ; and is, 
therefore, the point of intersection of the three such lines. 

Let ABC (Fig. 60) be any triangle, and D the middle point 
of BC. Then the centroid of ABC must lie on AD. For AD 
bisects all lines, such as be, parallel to BC. Now inscribe in 
the triangle any number of parallelograms such as bee'b', with 
sides parallel respectively to BC and 
AD. The centroid of each parallelo- 
gram lies on AD, and, therefore, so 
also does the centroid of the whole 
area composed of such parallelograms. 
If the number of such parallelograms 
be increased without limit, the alti- 
tude of each being diminished without limit, their combined 
area will approach that of the triangle, and the centroid of this 
area will approach in position that of the triangle. But since 
the former point is always on the line AD, its limiting position 
must be on that line. Therefore the line AD contains the cen- 
troid of the triangle. 

By the same reasoning, it follows that the centroid of ABC 
must lie on BE, drawn from B to the middle point of AC. 
Hence it must be the point of intersection of AD and BE, 
which point must also lie on the line CF drawn from C to the 
middle point of AB. 

The point G divides each bisector into segments which are 
to each other as i to 2. For, from the similar triangles ABC, 
EDC, since EC is half of AC, it follows that DE is equal to 
half of BA. And from the similar triangles AGB, DGE, since 
DE is half of AB, it follows that GE is half of GB, and GD 
half of GA. 

Qiiadrilateml. — Let ABCD (Fig. 61) be a quadrilateral of 
which it is required to find the. center of gravity. Draw BD, 
and let E be its middle point. Make EGi = \ EA, and EG% = \ 
EC. Then the centroids of the triangles ABD 3ind BCD are 



1 64 



GRAPHIC STATICS. 




G^ and G^ respectively. Hence the centroid of ABCD is on 
the line G^G^ at a point dividing it into segments inversely- 
proportional to the areas of ABD 
and BCD. Since these two tri- 
angles have a common base, their 
areas are proportional to their alti- 
tudes measured from this base. 
But these altitudes are propor- 
tional to AF and FC, or to G-^H and G^H ; hence, if G is the 
required centroid, 

G\G '. GiG : : G^H : G\H. 

Therefore G is found by making GiG=G2H. 

Circular sector. — To find the centroid of a circular sector 
OAB (Fig. 62) we may reason as follows : Draw two radii OM, 
ON, very near together. Then OMN differs little from a tri- 
angle, and its centroid will fall very near the arc A^B\ drawn 
with radius equal to two-thirds of OA. If the whole sector be 
subdivided into elements such as OMN, their centers of gravity 
will all fall very near to the arc AB\ If the 
number of such elements is indefinitely in- 
creased, the line joining their centroids 
approaches as a limit the arc A^BK And 
since the areas of the elements are propor- 
tional to the lengths of the corresponding 
portions of A^B\ the centroid of the total 
area is the same as that of the arc A^BK 
This point may be found by the method described in Art. 183. 

185. Graphic Determination of Areas. — Let there be given 
any number of geometrical figures, and let it be required to 
determine the relative magnitudes of their areas. 

If a number of rectangles can be found, of areas equal 
respectively to the areas of the given figures and having one 
common side, then the remaining sides will be proportional to 
the areas of the given figures. 



M N 




CENTROIDS OF LINES AND OF AREAS. 



165 



An important case is that in which the given figures are 
such that the area of each is equal to the product of two known 
lines. In this case a series of equivalent rectangles having one 
•common side can be found by the following construction. 

Let ABC (Fig. 63) be a triangle, and let it be required to 
determine an equivalent q 

rectangle having a side of ^ R 

given length as LM. Let 

■d and /i be the base and / 





altitude of the triangle. ^ ^ L n M 

MakeLN=^d and LP = /i, Fig.es 

and draw the semicircumference PRN. Draw the ordinate LR 
perpendicular to PN; then 

LR^ = PLx LN= 1 bh = area ABC. 

Draw MR, and perpendicular to it draw RQ. Then 

LR^=LMxLQ = 2ir&2iABC. 

Hence LQis, the required length. 

If the given figure is a parallelogram, LN and LP may be 
its' base and altitude. If it be a circular sector, LN and LP 
may be the length of the arc and half the radius. 

186. Centroids of Partial Areas. — It may be required to 
find the centroid of the part remaining after deducting known 
parts from a given area. For this case the construction of 
Art. 182 needs modification. In the case there considered we 
regarded the partial areas and the total area as proportional to 
parallel forces acting at their respective centers of gravity. In 
this case we may also represent the total area, the portions 
deducted from it, and the remaining portion as forces acting at 
the respective centers of gravity ; but the forces corresponding 
to the areas deducted must be taken as acting in the opposite 
direction to that assumed for the forces representing the total 
area and the area remaining. 

Thus, to find the centroid of the area remaining after deduct- 



i66 



GRAPHIC STATICS. 




ing from the circle ABD (Fig. 64) a smaller circle EFH and a 
sector OAB, we may proceed as follows : Find the centroid of 
three parallel forces proportional to the 
areas ABD, EFH, and OAB, and applied 
at their respective centroids O, C, and G\ 
but the last two must be taken as acting 
in the direction opposite to that of the first. 
With this understanding, the force and 
funicular polygons may be employed as in 
Fis.e4 Art. 172. 

187. Moments of Areas. — The moment of an area about any 
line in its plane may be determined from the funicular polygon 
employed in finding its center of gravity. Let the parallel 
forces applied at the centroids of the partial areas be assumed 
to act parallel to the axis of moments. Then the distance inter- 
cepted on the axis by the extreme lines of the funicular poly- 
gon, multiplied by the pole distance, is equal to the moment of 
the total area about the axis. For, by Art. 56, this construc- 
tion gives the moment of the resultant force about any point 
in the given axis ; and this is equal to the moment of the result- 
ant area about the axis by definition. 

A similar rule gives the moments of the partial areas. 



CHAPTER IX. MOMENTS OF INERTIA. 
§ I. Moments of Inertia of Forces. 

1 88. Definitions. — The moment of inertia of a body with 
respect to an axis is the sum of the products obtained by mul- 
tiplying the mass of every elementary portion of the body by 
the square of its distance from the given axis. 

The moment of inertia of an area with respect to an axis is 
the sum of the products obtained by multiplying each element- 
ary area by the square of its distance from the axis. 

The moment of inertia of a line may be similarly defined, 
using elements of length instead of elements of mass or area. 

The moment of inertia of a force with respect to any axis is 
the product of the magnitude of the force into the square of the 
distance of its point of application from the axis. The sum of 
such products for any system of parallel forces is the moment 
of inertia of the system with respect to the given axis. 

[Note. — The term moment of inertia had reference originally to material bodies, 
the quantity thus designated having especial significance in dynamical problems relat- 
ing to the rotation of rigid bodies. The quantity above defined as the moment of 
inertia of an area is of frequent occurrence in the discussion of beams, columns, and 
shafts in the mechanics of materials. In the graphic discussion of moments of iner- 
tia of areas, it is convenient to treat areas as forces, just as in the determination of 
centers of gravity; it is therefore convenient to use the term moment of inertia of a 
force in the sense above defined. It is only in the case of masses that the term 
moment of inertia is really appropriate, but it is by analogy convenient to apply it 
to the other cases.] 

The product of inertia of a mass with respect to two planes is 
the sum of the products obtained by multiplying each element- 
ary mass by the product of its distances from the two planes. 

167 



l68 GRAPHIC STATICS. 

The product of inertia of an area, a line, or a force may be 
defined in a similar manner. 

For a plane area, the product of inertia with respect to 
two lines in its plane may be defined as the sum of the 
products obtained by multiplying each element of area by the 
product of its distances from the given lines. This is equiv- 
alent to the product of inertia of the area with respect to two 
planes perpendicular to the area and containing the two given 
lines. 

For a system of forces with points of application in the same 
plane, the product of inertia with reference to two axes in that 
plane may be defined as the sum of the products obtained by 
multiplying the magnitude of each force by the product of the 
distances of its point of application from the given axes. It is 
with such systems and with plane areas that the following pages 
chiefly deal. 

The radius of gyration of a body with respect to an axis is the 
distance from the axis of a point at which, if the whole mass of 
the body were concentrated, its moment of inertia would be 
unchanged. The square of the radius of gyration is equal to 
the quotient obtained by dividing the moment of inertia of the 
body by its mass. 

The radius of gyration of an area may be defined in a similar 
manner. 

The radius of gyration of a system of parallel forces is 
the distance from the axis of the point at which a force 
equal in magnitude to their resultant must act in order that 
its moment of inertia may be the same as that of the system. 
The square of the radius of gyration may be found by dividing 
the moment of inertia of the system by the resultant of the 
forces. 

189. Algebraic Expressions for Moment and Product of In- 
ertia. — Motnent of inertia. — Let ^i, ni.^, etc., represent ele- 
mentary masses of a body, and ;i, 7% etc., their respective 



MOMENTS OF INERTIA OF FORCES. 169 

distances from an axis ; then the moment of inertia of the body 
with respect to that axis is 

the symbol % being a sign of summation, and the second mem- 
ber of the equation being merely an abbreviated expression for 
the first. 

Product of inertia. — Let/i, ^2, etc., denote the perpendicular 
distances of elements mx, lUi, etc., from one plane, and q^, q^, etc., 
their distances from another ; then the product of inertia of the 
body with respect to the two planes is 

inxpxqx + w/2/2^2 H = %mpq. 

Radius of gyration. — With the same notation, if k denotes 
the radius of gyration of the body, we have 

{iHx + m^ ■\ ) J^ = ni\r\ + w/^r./ -\ . 

Or, 

7 2 _ m^r^ + m.^r^ -\ _ S/wr^ 

nix + ^^^2 -\ S^^2 

Here '%m is equal to the whole mass of the body. 

Product-radius. — Let c represent a quantity defined by the 
equation. 

{vh + m2-\ ) ^ = wi/i^i + M2p2q2 -\ • 

Then if the whole mass were concentrated at the same distance 
c from two axes, its product of inertia with respect to those 
axes would be unchanged. This quantity c is thus seen to be 
analogous to the radius of gyration. It may be called the 
product-radius of the body with respect to the two axes. The 
value of c is always given by the equation 

Expressions similar to those just given apply also to plane 
areas and to systems of parallel forces. In case of an area, 
'Ml, mi, etc., denote elements of area ; r^, ri, etc., their distances 
from the axis of inertia ; and (/i, q^, {p^, q^, etc., their dis- 



I/O 



GRAPHIC STATICS. 



tances from the two planes. In case of a system of parallel 
forces with complanar points of application, Wi, m^, etc., must 
be replaced by the magnitudes of the forces, 

190. Determination of Moment of Inertia of a System of 
Parallel Forces. — Let the points of application of the forces 
be in the same plane, which also contains the assumed axis. 
We shall have to deal only with systems satisfying these condi- 
tions. By the definition, the moment of inertia will be the 
same, whatever the direction of the forces. If we take the 
moment of any force (as defined in Art. 175) about the given 
axis, and suppose a force equal in magnitude to this moment 
to act at the point of application of the original force, and in a 
direction corresponding to the sign of the moment, then the 
moment of this new force about the given axis is equal to 
the moment of inertia of the original force. If this be done for 
all the forces, the algebraic sum of the results will be the 
required moment of inertia of the system. 

This process can be carried out graphically by methods 
already described. 

Let ab, be, cd, de (Fig. 65) be the points of application of 
four parallel forces, and let the axis of inertia be QR. Suppose 
all the forces to act in lines parallel to QR, passing through the 
given points of application. Their respective moments with 
reference to QR may now be found by the method of Art. 55. 

Draw the force polygon ABCDE and choose a pole O, taking 
the pole distance H preferably equal to AE or some simple 
multiple of AE. (In Fig. 65 H is taken equal to AE) Draw 
a funicular polygon and prolong each string to intersect QR. 
Then the moment of any force with respect to QR is the 
product of H by the distance intercepted on QR by the two 
strings corresponding to the force in question. Thus the 
moment of AB is given by the distance A' B' (Fig. 65) multi- 
plied by H. Also the successive moments of BC, CD, DE are 
represented by B^C\ CD' , D' E' , each multiplied by H. It is 



MOMENTS OF INERTIA OF FORCES. 



171 



seen also that the intercepts, if read in the above order, give a 
distinction between positive and negative moments ; upward 
distances on QR denoting in this case positive moments, and 
downward distances negative moments. 

We have now to find the sum of the moments of a second 
system of forces acting in the original lines, but represented in 





Fig. 65 



magnitude and direction by the intercepts just found. We 
may take as the force polygon for the second construction the 
line A'B' C'D'E' (Fig. 65), and choosing any pole distance H', 
draw a second funicular polygon and find the distances inter- 
cepted by the successive strings on the line QR. But in this 
case, since only the resultant is desired, we need only find the 
intercept A" E" between the first and last strings. The product 
of this intercept by H' gives the sum of the moments of A'B', 
B'C, CD', D'E' with respect to QR ; and, if the product be 
multiplied by H (since A' B' , B'C, etc., should each be multi- 
plied by H in order to represent the magnitudes of the forces 
of the second system), the result will be the required moment of 
inertia of the given system of forces AB, BC, CD, DE. 



172 GRAPHIC STATICS. 

It should be noticed that in Fig. 65 (A) is a force diagram, 
{B) a space diagram (Art. 11); that is, every line in {A) repre- 
sents a force, while every line in {E) represents a distance. 
Even A' B' , B' C , etc., though used as forces, are actually 
merely distances ; and the moment of any one of them is the 
product of a length by a length. 

191. Radius of Gyration. — The moment of inertia of the 
given system is HxH' xA" E" . If H has been taken equal 
to AE, the product H'xA"E" must equal the square of the 
radius of gyration of the system with respect to QR. The 
length of the radius of gyration can be found as follows (Fig. 
65) : Draw LN=H' -\-A" E" , and make LM==H'. On ZTV^as 
a diameter draw a semicircle LPN, and from M draw a line 
perpendicular to LN, intersecting the semicircle in P. Then 
MP is the length of the required radius of gyration. For by 
elementary geometry we have PM =LMxMN. 

If His taken equal to nxAE, the moment of inertia is equal 
to H' xA"E" X ;/ x AE, and the square of the radius of gyration 
is equal to fiH' xA"E". Hence in Fig. 65 we should put 
either LM=nH\ or MN=nxA"E". 

192. Central Axis. — If the axis with reference to which the 
moment of inertia is found contains the centroid of the given 
system of forces, it is called a central axis of the system. 

In many cases it is desired to find the moment of inertia with 
respect to a central axis whose direction is known while the 
position of the centroid is at first unknown. It is to be noticed 
that the method shown in Fig. 65 is applicable in this case ; 
for the first part of the process is identical with that employed 
in finding the centroid of the system. If, in Fig. 65, the 
strings oa, oe, of the first funicular polygon be prolonged to 
intersect, a line through their point of intersection, parallel to 
the direction of the forces, will contain the centroid of the 
system. If this line is taken as the inertia-axis, the points A' , 
B' , C , D' , E' are the points in which this axis is intersected by 



MOMENTS OF INERTIA OF FORCES. 



173 



the strings oa, ob, oc, od, oe. No further modification of the 
process is necessary. 

193. Moment of Inertia Determined from Area of Funicular 
Polygon. — In Fig. 65, the moments of the given forces are 
represented by the intercepts A^ B\ B' C , CD', D'E', each 
multiplied by the pole distance H. The moment of inertia of 
any force, as AB, is equal to the moment of a force represented 
by the corresponding intercept as A' B' , supposed to act in the 
line ab. Now, by definition (Art. 175) the moment about the 
axis QR of a force equal to A'B' acting in the line ab is equal 
to double the area of the triangle A' i B' ; hence the moment of 
the force HxA'B' is equal to double the area of that triangle 
multiplied hy H. Similarly, the moment of a force HxB'C, 
acting in the line be, is equal to double the area of the triangle 
B' 2 C multiplied by H. Applying the same reasoning to each 
force, we see that the sum of the moments of the assumed 
forces {HxA'B', HxB'C, etc.) is equal to 2 /z*" times the sum 
of the areas of the triangles A' i B', B' 2 C, C 3 D' , D' 4E'. 
In adding these triangles, each must be taken with its proper 
sign, corresponding to the sign of the moment represented by 
it. Thus, the moments of A'B' , B' C, and D'E' all have the 
same sign, while the moment of CD' has the opposite sign. 
The algebraic sum of the areas is, therefore, 

area A' i ^' + area B' 2 C — area C 3 Z>' + area D' 4E' , 

which is equal to the area of the polygon A' i 2 34E'. Hence 
this area, multiplied by 2 //, gives the moment of inertia of the 
required system of forces. 

It may sometimes be convenient to apply this principle in 
determining moments of inertia, the area being determined by 
use of a planimeter, or by any other convenient method. It 
should be noticed that if H is taken equal to the sum of the 
given forces (AE), twice the area of the funicular polygon is 
equal to the square of the radius of gyration. If -fl=l- AE, the 



174 



GRAPHIC STATICS. 



square of the radius of gyration is equal to the area of the 
polygon. 

194. Determination of Product of Inertia of Parallel Forces. 

— Assume the points of application of the forces to be in the 
plane containing the two axes. If the moment of any force 
with respect to one axis be found, and a force equal in magni- 
tude to this moment be assumed to act at the point of applica- 
tion of the original force, then the moment of this new force 
with respect to the second axis is equal to the product of 
inertia of the given force for the two axes. 

Thus, let ab, be, cd, de (Fig. 66) be the points of application 
of four parallel forces, and let their product of inertia with 
respect to the axes QR, ST be required. Draw ABCDE, the 




Fig. 66 



force polygon for the given forces, assumed to act parallel to 
QR. Choose a pole O, the pole distance being preferably 
taken equal to AE, or some simple multiple of AE, and draw 
the funicular polygon as shown, prolonging the strings to inter- 
sect QR in the points A\ B\ C, D\ E'. Now assume a series 



MOMENTS OF INERTIA OF FORCES. 175 

of forces equal to A'B\ B'C\ etc., each multiplied by H, to act 
at the points ab, be, etc., and determine their moments with 
respect to the axis ST. To find these moments, draw lines 
through ab, be, etc., parallel to ST, and draw a funicular poly- 
gon for the assumed forces taken to act in these lines. The 
force polygon is obtained by revolving the line A'B'C'D'E' until 
parallel with ST, and is the line A\ B\ C\ D\ E\ in the figure. 
The strings o^a\ o'e' of the second funicular polygon intersect 
ST in the points A" and E". Hence, calling //' the second 
pole distance, the product of inertia of the given system is 
equal to i/xi/'x^"^". 

195. Product-Radius. — If Hhe taken equal to AE (Fig. 66), 
H' xA"E" is equal to the square of the product-radius (Art. 
189). Hence the product-radius can be found by a construc- 
tion exactly like that employed in finding the radius of gyra- 
tion. Thus (Fig. 66) take LM=H' and AIN=A"E"; make 
LN the diameter of a semicircle, and draw from M a line per- 
pendicular to LN, intersecting the semicircle in P. Then 
'Mp'=LMxAfN=H'xA''E''; hence MP = e, the product- 
radius. 

196. Relation between Moments of Inertia for Parallel Axes. 
— Proposition. — The moment of inertia of a system of par- 
allel forces with reference to any axis is equal to its moment of 
inertia with respect to a parallel axis through the centroid of 
the system plus the moment of inertia with respect to the given 
axis of the resultant applied at the centroid of the system. 

Let P\, P2, etc., represent the forces ; Xi, Xo, etc., the dis- 
tances of their points of application from the central axis ; and 
a the distance of this central axis from the given axis. Calling 
the required moment of inertia A, and the moment of inertia 
with respect to the axis through the center of gravity A', we 
have 

A=P,{a+x,Y + P,{a+^^y+- 

= « 2 (^1 + ^2 + • • •) + 2« (Pj-ri -f P2X2 +••■)+ Pi^rr + P.xi + • • • 



1^6 GRAPHIC STATICS. 

Now PJX1 + P2X2-] — is the algebraic sum of the moments of 
the given forces with respect to the axis through their centroid, 
and is equal to zero ; and P^Vi^ + Pz^^^^ — =A'. Hence 

A = A'+{P^ + P2+'")a', 

which proves the proposition. 

Radii of gyration. — Let k = radius of gyration of the sys- 
tem with respect to the given axis, and k^ the radius of gyra- 
tion with respect to the central axis, and we have 

A = {P^^P,^-)k\ 

A'={P, + P2+-)k'\ 

Hence the equation above deduced may be reduced to the form 

197. Products of Inertia with Respect to Parallel Axes. — 

Proposition. — The product of inertia of a system of parallel 
forces with reference to any two axes is equal to the product 
of inertia with reference to a pair of central axes parallel to the 
given axes, plus the product of inertia of the resultant (acting 
at the centroid) with reference to the given axes. 

Let P^, P2, etc., be the magnitudes of the given forces ; {pi, g^,. 
{p2, q-i), etc., the distances of their points of application from 
the central axes parallel to the two given axes ; {a, b) the dis- 
tances of the centroid of the system from the given axes. Let 
A and A' be the products of inertia of the system with respect 
to the given axes and the parallel central axes respectively. 

Then 

A' = P,Pig^ + P2p.q2+"- 

A =P,{pi + a){q, + b) +P2{p2+a){q2 + b) + "- 

= {Pip^qi + P2p2q2 +■•')+ a {P^q^ + P2q2 +"■) 

+ b{P,p, + P2P2+-)-\-ab{P^-\-P2+'-). 

But P\q\-\-P2q2-\ — =0; and Pipi + P2p2-\ — = 0; since each of 
these expressions is the sum of the moments of the given forces 



MOMENTS OF INERTIA OF PLANE AREAS 177 

with respect to an axis through the centroid of the system. 

Hence, 

A = A'+ {P, + P2+"-)ab, 

which proves the proposition. 

liA = {P, + P^-\--) c" and A^={P^J^P^-^...y\ we have 

C' = c +ao. 

From the above proposition it follows that if the axes have 
such directions that the product of inertia with reference to the 
central axes is zero, the product of inertia with reference to the 
given axes is the same as if the forces all acted at the centroid. 
When this condition is known to be satisfied, then for the pur- 
pose of finding the product of inertia the system of forces may 
be replaced by their resultant. 

It follows also, in the case when the product of inertia for the 
central axes is zero, that if one of the given axes coincides with 
the parallel central axis, the product of inertia for the given 
axes is zero ; for in this case either a or b is zero, and hence 
ab{Pi-\-P2-\ — ) is zero. Therefore, 

If the product of inertia of a system is zero for two axes, A' 
and A" , one of which (as A') contains the centroid of a system, 
then the product of inertia is also zero for A' and any axis 
parallel to ^". 

§ 2. Moments of Inertia of Plane Areas, 

198. Elementary Areas Treated as Forces. — If any area be 
divided into small elements, and a force be applied at the 
centroid of each element numerically equal to its area, the 
moment of inertia of this system of forces will be approximately 
equal to that of the given area. The approximation will be 
closer the smaller the elementary areas are taken. If the ele- 
ments be made smaller and smaller, so that the area of each 
approaches zero as a limit, the moment of inertia of the sup- 
posed system of forces approaches as a limit the true value of 
the moment of inertia of the given area. 



1/8 GRAPHIC STATICS. 

It is seen, then, that most of the general principles which 
have been stated regarding moments of inertia of systems of 
forces are equally applicable to moments of inertia of areas. 
The practical application of these principles, however, and 
especially the graphic constructions based upon them, are less 
simple in the case of areas than of systems of forces such as 
those already treated. The reason for this is that the system 
of forces which may be conceived to replace the elements of 
area consists of an infinite number of infinitely small forces, 
with which the graphic methods thus far discussed cannot 
readily deal. Problems of this class are most easily treated by 
means of the integral calculus, especially when the areas dealt 
with are in the form of geometrical figures. It is possible, 
however, by graphic methods to determine approximately the 
moment of inertia of any plane area ; and in many cases exact 
graphic solutions of such problems are not difficult. The proof 
of these methods is often most easily effected algebraically. 

199. Moments of Inertia of Geometrical Figures. — The appli- 
cation of the integral calculus to the determination of moments 
of inertia will not be here treated. But the values of the 
moments of inertia of some of the common geometrical figures 
are of such frequent use that the more important of them will 
be given for future reference. The moment of inertia is in each 
case taken with respect to a central axis, and will be repre- 
sented by /, while the radius of gyration will be called k. 

Rectangle. — Let b and d be the sides, the axis being parallel 
to the side b. Then 

12 ' 12 

Triangle. — Let b and d be the base and altitude. Then for 
an axis parallel to the base, 

36 ' i8' 



MOMENTS OF INERTIA OF PLANE AREAS. 179 

For an axis through the vertex, bisecting the base, k'^ = — , 

where b' is the projection of the base on a Hne perpendicular to 
the axis. 

Circle. — Let d be the diameter. Then 

64 ' 16 

For a central axis perpendicular to the plane of the circle, 

1='^^- B = — 
32 ' 8' 

Ellipse. — Let a and b be the semi-axes. Then for an axis 
parallel to a, 

j_ rrrab^ . h'^ — — 
4 ' 4 

For an axis parallel to b, 

j_ ira^b . jji_a^ 
4 ' 4 

For a central axis perpendicular to the plane of the ellipse, 

4 ' 4 

Graphic construction for radius of gyration. — Whenever k'^ 
can be expressed as the product of two known factors, the value 

of k can be found by the construction already used in Art. 191. 

^2 
Thus, in case of a rectangle, for which li^ = — , we may put 

12 

j^^^d d ^^^^ .^ .^ p. g ^^ ^^^^ LM=~, MN=-, the 

3 4 4 3 

construction there shown will give MP as the value of k. For 

the triangle, the axis being parallel to the base, we have 

k"^ = - ■ -, and the same construction is applicable. For the 
3 6 
. . b' b' 

axis through the vertex bisecting the base, /^^ = — • —• 

4 o 



l8o GRAPHIC STATICS. 

200. Product of Inertia. — General principles. — Products of 
inertia of areas are determined by means of the integral calculus 
in a manner similar to that employed for moments of inertia. 
The following fundamental principles regarding products of 
inertia of geometrical figures will be found useful : 

(i) With reference to two rectmigiUar axes, one of which is 
an axis of symmetry (Art. 179), the product of inertia is zero. 
For it is manifest that the products of inertia of two equal 
elements, symmetrically placed with reference to one of the 
axes, are numerically equal but of opposite sign. Hence, if 
the whole area can be made up of such pairs of elements, the 
total product of inertia is zero. 

(2) If the two axes are not rectangular, but the area can be 
divided into elements such that for every element whose dis- 
tances from the axes are p, q, there is an equal element whose 
distances are /, —q, or — /, q, the product of inertia is zero. 
This includes the preceding as a special case. 

201. Products of Inertia of Geometrical Figures. — In each of 
the following cases the product of inertia is zero : 

A triangle, one axis containing the vertex and the middle 
point of the base, the other being any line parallel to the base. 

A parallelogram, the axes being parallel to the sides and one 
axis being central. This includes the rectangle as a special 
case. 

An ellipse, the axes being parallel to a pair of conjugate 
diameters, and one axis being central. This includes, as a 
special case, that in which one axis is a principal diameter and 
the other is any line perpendicular to it ; and under this case 
falls also the circle. 

202. Approximate Method for Finding Moment of Inertia of 
Any Area. — To apply the method of Art. 190 to the determina- 
tion of the moment of inertia of a plane area, we should strictly 
need to replace the area by an infinite number of parallel forces, 
proportional to the infinitesimal elements of the given area, and 



MOMENTS OF INERTIA OF PLANE AREAS. 



I8l 



with points of application in these elements. If, instead, we 
divide the given figure into finite portions whose several areas 
are known, and assume forces proportional to those areas to 
act at their centroids, we may get an approximate value for the 
moment of inertia, which will be more nearly correct the smaller 
the elements. This will be illustrated by the area shown in 
Fig. 67. 

Let QR be the axis with reference to which the moment of 
inertia is to be found, in this case taken as a central axis. 




Divide the figure into four rectangular areas as shown, and 
assume forces numerically equal to these areas to act at their 
centroids parallel to QR. The force polygon is ABODE. 
Draw the funicular polygon corresponding to a pole O, and let 
the successive strings intersect QR in A', B', C , D' , E'. Take 
this as a new force polygon, and with any convenient pole 
distance draw a second funicular polygon, using the same lines 
of action. Let the first and last strings intersect QR in A" 
and E". Then A" E" multiplied by the product of the two 
pole distances gives the moment of inertia of the four assumed 
forces, and approximately the moment of inertia of the given 



l82 GRAPHIC STATICS. 

figure. If the first pole distance is taken equal to AE (as is. 
the case in Fig. Oy), the radius of gyration may be found by the 
construction of Art. 191. Thus in Fig. 6"/, MP is the radius 
of gyration as determined by this method. 

A more accurate result may be reached by dividing the area 
into narrower strips by lines parallel to QR ; since the narrower 
such a strip is, the more nearly will the distance of each small 
element from the axis coincide with that of the centroid of the 
strip. If the partial areas are taken as narrow strips of equal 
width, the forces may be taken proportional to the average 
lengths of the several strips. 

203. Accurate Methods. — If the given figure can be divided 
into parts, such that the area of each is known, and also its 
radius of gyration with respect to its central axis parallel to the 
given axis, the above method may be so modified as to give an 
accurate result. Two methods will be noticed. 

(i) When the axis is known at the start. — Let the line of 
action of the force representing any partial area be taken at a 
distance from the given axis equal to the radius of gyration of 
that area with reference to the axis. If this is done, it is 
evident that the moment of inertia of the system of forces is 
identical with that of the given area. When the axis is known, 
the position of the line of action for any force may be found as 
follows : Let QR (Fig. 6%) be the given axis, 
and Q R^ a parallel axis through the centroid 
of any partial area. Draw KL perpendicu- 
lar to QR, and lay off LM equal to the 
radius of gyration of the partial area with 
68 " respect to Q R' . Then KM is the length 
of the radius of gyration with respect to QR. (Art. 196.) Take 
iTiV equal to KM and draw 0"^" through N parallel to Q' R^ ; 
then Q"R" is to be taken as the line of action of the force 
representing the partial area in question. 

(2) When the axis is at first unknown. — The method to be 




MOMENTS OF INERTIA OF PLANE AREAS. 



183 



employed in this case is to let the force representing any partial 
area act in a line through the centroid of that area ; and then 
assume the force representing its moment to act in such a line 
that the moment of this second force shall be numerically 
equal to the moment of inertia of the partial area. This line 
may be found as follows : Let k represent the radius of gyra- 
tion of the partial area with respect to its central axis parallel 
to the given axis, and a the distance between the two axes. 
Then the moment of inertia of the partial area with respect 
to the given axis is A {cP'-\-k'^), if A represents the area. But 
A (a'^-irk^)=Aa(a-\ — J. Hence, if a force numerically equal to 



A is assumed to act with an arm a, then a force equal to its 

P . 
moment Aa must act with an arm a-\ — in order that its moment 

a 
may equal A (a^-{-k^). 

The distance a-\ — can be found by a simple construction.. 
a 

Let QR and Q'R' (Fig. 69) be the given axis and the central 

axis respectively. Draw KL 

perpendicular to QR and lay off 

LM equal to k. From 31 draw 

a line perpendicular to KM, in- j^i/ ^ ^^1 Fig. 69 

tersecting KL produced at JV. 

Then KJV=a+-- For, in the 

a 
similar triangles KNM, KML, we have 



q' 


/ 


Q' 
M 


Q 


N 


/ 


L ^^ 


K 


R 




R' 


R 



KN ^KM , 
KM KL ' 



or KJV= 



KM^ 
KL ' 



But KL=a, and KM^ = P + P ; hence 

/72 4. A2 Z,2 



This second method is more useful than the first, because in 
applying it the first funicular polygon can be drawn before the 
position of the inertia-axis is known. Thus, a very common 



1 84 



GRAPHIC STATICS. 



case is that in which the moment of inertia of an area is to be 
found for a central axis, whose direction is known, while at the 
outset its position is unknown because the centroid of the area 
is unknown. If the second method be employed, the first 
funicular polygon can be drawn at once, and serves to locate 
the required central axis, as well as to determine the moments 
of the first set of forces as soon as the axis is known. The 
central axis and the moment of inertia with respect to it are 
thus determined by a single construction. 

Example. — The method last described is illustrated in Fig. 
70. The area shown consists of two rectangles, the centroids 
of which are marked ab and be. The moment of inertia is to 
be found for a central axis parallel to the longer side of the 
rectangle ab. 




We draw through ab and be lines parallel fo the assumed 
direction of the axis, and take these for the lines of action of 
forces AB, BC, proportional to the areas of the two rectangles. 
ABC is the force polygon, and the pole distance is taken equal 



MOMENTS OF INERTIA OF PLANE AREAS. 185 

to AC. The intersection of the strings oa, oc, determines a 
point of the required central axis ac. The moments of the 
given forces are proportional to A'B', B'C. We now have to 
find the lines of action for the forces A'B', B'C, in accordance 
with the method above described. 

Take any line perpendicular to the axis ac, as VQ, the side 
of one of the rectangles. From R, the point in which this line 
intersects the vertical line through be, lay oH RT equal to the 
central radius of gyration of the rectangle whose centroid is 
be. To find this central radius of gyration, we know that its 

value is a/ — (Art. 199), where d is the length of the side of 
the rectangle perpendicular to the axis. Hence we take 

QR = -, RS=—, and make QS the diameter of a semicircle, 

3 4 

intersecting the vertical line be in T; then RT is the required 
radius of gyration of the rectangle with respect to a central 
axis. Now draw from T a line perpendicular to VT, intersect- 
ing VQ in U; then the line b'e' drawn through f/ parallel to the 
given axis is the line of action of the force B'C. 

By a similar construction applied to the other rectangle, a'b' 
is located as the line of action of the force A'B'. The second 
funicular polygon is now drawn, and the points A", C" are 
found by the intersection of the strings o'a', o'e' with the axis. 
Hence the moment of inertia of the given area is equal to 
A"C" xHxH'. In the figure H is made equal to AC, hence 
the radius of gyration can be determined by the usual con- 
struction, and its length is found to be MP. 

204. Moment of Inertia of Area Determined from Area of 
Funicular Polygon. — The method given in Art. 193 for finding 
the moment of inertia of a system of forces by means of the 
area of the funicular polygon may be applied with approximate 
results to the case of a plane figure. If the forces are taken 
as acting at the centroids of the areas they represent, then to 



l86 GRAPHIC STATICS. 

get good results these partial areas should be taken as narrow 
strips between lines parallel to the axis. 

If the lines of action are determined as in the first method 
of the preceding article, then the area enclosed by the funic- 
ular polygon and the axis represents accurately the required 
moment of inertia. 

205. Product of Inertia Determined Graphically. — To deter- 
mine the product of inertia of any area, let it be divided into 
small known parts, and let parallel forces numerically equal to 
the partial areas be assumed to act at the centroids of these 
parts. The product of inertia of these forces may then be 
found as in Art. 194, and its value will represent approximately 
the product of inertia of the given area. 

If the partial areas can be so taken that the product of 
inertia of each with reference to axes through its centroid par- 
allel to the given axes is zero, the method here given is exact. 
(Art. 197.) 

If the partial areas are taken as narrow strips parallel to one 
of the axes, the condition just mentioned will be nearly fulfilled ; 
for the product of inertia of each strip for a pair of axes 
through its centroid, one of which is parallel to its length, will 
be very small. 



CHAPTER X. CURVES OF INERTIA. 



General Prmciples. 



206. Relation between Moments of Inertia for Different Axes 
through the Same Point. — The moment of inertia with respect 
to any axis through a given point can be expressed in terms of 
the moments and product of inertia for any two axes through 
that point. It is necessary here to use algebraic methods, but 
the results reached form the basis of graphic constructions. 

Let OX, 6>F(Fig. 71) be the two given axes; 6, the angle 
included between them ; P^, P^, etc., the forces of the system ; x', 
y' , the coordinates of the point 
of application of any force P, 
referred to the axes OX, OY; 
p, q, the perpendicular distances 
of the same point from O Y and 
6>Jf respectively, so that 

p=x^ ^\wO, q=y sinO. 




Let a', b\ and c^ be quantities defined by the equations 



tp 



P+P2+ 

12 



XP 



^,,^ tPxy 



XP 



187 



1 88 GRAPHIC STATICS. 

Then 



^d — — — = ^ ^ > 



""^^"- %P XP 

^„^^,Q^tPy-j\n-e^tP£^ 

tp %p 

c'^ sm^ ^ = ^ =.-—£-1 ; 

tP SP 

and rt' sin Q, b' sin ^, c' sin ^ are respectively the radius of 
gyration with respect to O Y, the radius of gyration with respect 
to OX, and the product-radius (Art. 189) with respect to OX 
and OY. 

The moment of inertia (/) and radius of gyration {k) of the 
system for the axis OM, making an angle j) with OX, may now 
be computed as follows : 

Let s = perpendicular distance of the point of application of 
any force P from OM. Then from the geometry of the figure 
it is seen that 

s=y' sin (O — cp) —x' sin cj). 
Hence 

/= tPs'^=tPy^ sin2 (^ -(/,)_ 2 tPx'y' sin {6 - 4>) sin <f> 
+ tPx'^ sin^ (}i ; 

I^V^tP • sin2 (e-4,)-2c'^tP • sin (6'-(/)) sin (f) 

the factors involving 6 and ^ being constant for all terms of 
the summation. Hence 

P = ^ = d'^sm^{6-(P)-2c'^sm{e-(f))smcf) + a'^sm^(f> . . (i) 

From these equations / and k may be computed if a', b\ and 
c^ are known ; that is, if the moments and product of inertia for 
the two axes OX and O Y are known. 

Special case. — If ^ = 90°, the equation (i) becomes 

^2__^'2 (,Qg2^_2^'2 gjj^^ (;.QS ^_|_^'2 gij-^2^ (2) 



GENERAL PRINCIPLES. 1 89 

207. Products of Inertia for Different Axes through the Ori- 
gin. — The product of inertia with respect to OM and OX may 
be found as follows : 

Let A = the required product of inertia ; then 

A = tP^s=tPy sin 6* [/ sin {d-(f))-x' sin </>] 
= tP [ j'2 sin sin (0 - </>) -;r>' sin 6 sin </>] 
= [5'2 sin e sin (6' - </>) - c'^ sin 6' sin cj)] tP. 

Let /^ = product-radius for axes OJ/and (^Jf; then 

h'^=i—-- = b''^ sin ^ sin (O — cfi) ~c'^ sin ^ sin 0. 

Special case. — The axis OM may be so chosen that A=o. 
This will be the case if 

^'2 sm{e-(j))=c''^ sin0. 

208. Inertia Curve. — If on OM (Fig. 71) a point Mhe taken 
such that the length (9 Jf depends in some given way upon the 
value of k, and if similar points be located for all possible direc- 
tions of OM, the locus of such points will be a curve which is 
called a curve of inertia of the system for the center O. 

The form of the curve will depend upon the assumed law 
connecting OM with k. 

209. Ellipse or Hyperbola of Inertia. ■ — The simplest curve is 
obtained by assuming OM to be inversely proportional to k. 

Let OM=r, and take r^= — j where d'^ is a positive 

quantity, so that d always represents a real length, positive or 
negative. 

Equation (i) of Art. 206 then becomes 

d^ sin2 6' = ^'V2 sin^ (61 - <jf>) - 2 ^'V^ sin {6 - (/,) sin + «'V2 sin^ ^, 

which is the polar equation of the inertia-curve, r and being 
the variable coordinates. Let x, y be coordinates of the point 
M referred to the axes OX, O Y. Then 

r _ X _ y 
' sin ^ sin(^ — (^) sin 



IQO GRAPHIC STATICS. 

whence r4 ^^=^, 

siri'^ 6 

sin'^ t' 

jT^ sin (^ — 0) sin _ 
sm"' 6 
and the equation becomes 

^4^^^2^,2_2^'2_^_y4_^'2y (3) 

The form of this equation shows that it represents a conic 
section whose center is at the origin of coordinates O. This 
conic may be either an elHpse or a hyperbola. 

The equation will be discussed more fully in a later article. 
One fact may, however, be here noticed. 

If the moment of inertia / is positive for all positions of the 
axis, the radius of gyration k will be real, whatever the value 
of ^. But r, the radius vector of the curve, will be real when k 
is real ; hence in this case the curve is an ellipse. This is 
always the case if the given forces have all the same sign. 

If the forces have not all the same sign, it. is possible that 
the value of / may have different signs for different directions 
of the axis. If this is so, certain values of ^ make k (and 
therefore r) imaginary. In this case the curve is a hyperbola. 

The most important case is that in which the given forces 
have all the same sign which may be taken as plus, so that the 
moment of inertia is always positive, and the curve an ellipse ; 
and to this case the discussion will be confined. 

§ 2. Inertia-Ellipses for Systetns of Forces. 

2IO. Properties of the Ellipse. — In discussing ellipses of in- 
ertia use will be made of certain general properties of the 
ellipse, which, for convenience of reference, will be here sum- 
marized. For the proof of the propositions stated the reader 
is referred to works on the conic sections. 



INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. jgi 
(i) The equation 

represents an ellipse if Br' — AC is negative ; a hyperbola if 
B'^ — AC is positive. (Salmon's Conic Sections, p. 140.) The 
coordinate axes may be either rectangular or oblique. 

(2) Two diameters of an ellipse are said to be conjugate to 
each other if each bisects all chords parallel to the other. If 
the axes of coordinates coincide with a pair of conjugate 
diameters, the lengths of which are 2 a' and 2 b' , the equation 
of the curve is 

A particular case of this equation is that in which the coordi- 
nate axes are rectangular, being the principal axes of the curve ; 
in which case we may write a and b instead of a' and b'. 

(3) In an ellipse, the product of any semi-diameter and the 
perpendicular from the center on the tangent parallel to that 
semi-diameter is constant and equal to ab. That is, if r is any 
radius vector of the curve drawn from the center, and p the 
length of the perpendicular from the center to the parallel 
tangent, we have 

pr=ab 

where a and b are the principal semi-axes of the curve. 

(4) Let a' and b' be conjugate semi-diameters. Then each is 
parallel to the tangent at the extremity of the other. Hence 
the length of the perpendicular from the center to the tangent 
parallel to a' is b' sin 9, where 9 is the angle included between 
.a' and b' . Therefore from the preceding paragraph, 

a'b' sin 6 = ab. 

(5) An ellipse can be constructed, when a pair of conjugate 
diameters is known, as follows : 




192 GRAPHIC STATICS. 

Let AA', BB' (Fig. 72), be the conjugate diameters, O being 
the center of the ellipse. Complete the parallelogram OBCA. 

Divide OA and CA into parts 
proportional to each other, be- 
ginning at O and C. Through 
the points of division of OA 
draw lines radiating from B' , 
and through the points of divi- 
sion of CA draw lines radiating 
from B. The points of inter- 
section of the corresponding lines in the two sets are points of 
the ellipse. In a similar way, the other three quadrants may 
be drawn. (The location of one point is shown in the figure.) 

A convenient way to locate the corresponding points of 
division on OA and CA is to cut these lines by lines parallel tO' 
the diagonal OC 

211. Discussion of Equation of Inertia-Curve. — We will now 
examine the equation of the inertia-curve, 

with reference to the properties of the ellipse above enumerated. 

(i) If a'^b''^ — c''^ is positive, the equation denotes an ellipse. 
This cannot be the case if a'"^ and b'"^ have opposite signs. But 
from the definitions of a'"^ and b'^ (Art. 206) it is seen that their 
signs are the same as those of the moments of inertia for Fand 
X axes respectively. Hence, if there are any two axes through 
the assumed center for which the moments of inertia have 
opposite signs, the inertia-curve is a hyperbola. 

If the moment of inertia has the same sign for all axes 
through the assumed center, the curve is an ellipse. For, since 
c'"^ sin^d = — =r^ (Art. 206), <:' may be made zero by choosing the 

axes so that the product of inertia with respect to them is 
zero ; and if c^ is zero, and «'^ and V^ have the same sign, the 
quantity a'^b'^ — c'^ is positive. 



INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 193 

This agrees with the conclusion stated in Art. 209. 
We shall here deal only with ellipses of inertia. 

{2) If c' = o, the coordinate axes are conjugate axes of the 
curve. But the condition c' = o means that the product of 
inertia for the two axes is zero. Hence any two axes for which 
the product of inertia is zero are conjugate axes of the inertia- 
curve. (This is true whether the curve is an ellipse or a 
hyperbola.) 

(3) By the law of formation of the inertia-conic (Art. 209), 
the length of the radius vector lying in any line is inversely 
proportional to the radius of gyration with respect to that line. 
But by (3) of the last article, the perpendicular from the center 
on the tangent parallel to any radius vector is inversely propor- 
tional to the length of that radius vector. Hence the perpen- 
dicular distance between any diameter and the parallel tangent 
is directly proportional to the radius of gyration with respect to 
that diameter. The curve may be so constructed that the 
length of this perpendicular is e^ual to the radius of gyration, 
as follows : 

From Art. 209, we have 

7 _ <3^^ sin 
r 

and from (3) and (4) of the last article we have 

._ab_ a^b^ sin Q 
r r 

if a' and b^ are conjugate semi-diameters. Now take 

d''' sin 6 = ab = a'b' sin 6, 

or d'' = a'b', 

and we have k=p, 

and the equation of the curve becomes (since c' = o when the 
axes are conjugate) 



194 



GRAPHIC STATICS. 



If the equation be written in this form, a! and b^ having the 
meanings assigned in Art. 206, the radius of gyration about any 
axis throiigJi the center of the ellipse is equal to the perpe^idiadar 
distance between the axis and the parallel tangent to the ellipse. 

Hereafter we shall mean by inertia-ellipse the curve obtained 
by taking d^ = a'b' as above described, so that the radius of gyra- 
tion for any axis can be found by direct measurement when a 
parallel tangent to the ellipse is known. 

212. To Determine Tangents to the Inertia-Ellipse for Any- 
Center. — Let the radius of gyration {k) be found for any 
assumed axis through the given center by one of the methods 
already described (Arts. 202 and 203). Then two lines par- 
allel to the axis and distant k from it, on opposite sides, will be 
tangents to the inertia-ellipse. 

213. To Construct the Inertia-Ellipse, a Pair of Conjugate 
Axes Being Known in Position. — If the positions of two axes 
conjugate to each other can be found, the ellipse can be drawn 
by the following method : 

Determine the radius of gyration for each of the two axes 
and draw the corresponding tangents as in the preceding 
article ; then proceed as follows : 

Let XX', VV (Fig. 73) be the given axes, and let the four 
tangents determined as above form the parallelogram PQRS. 

Let A, A', B, B' be the points 
in which these tangents in- 
tersect the axes XX', YV. 
Then, since each diameter is 
parallel to the tangents at the 
extremities of the conjugate 
diameter, A, A', B, B' are the 

The 

elhpse can now be constructed as explained in Art. 210 
(Fig, 72). 

This method of constructing the inertia-ellipse is useful 




B/ R 

-^Y' Fig. T'S 

extremities of the diameters lying in the given axes 



INERTIA-CURVES FOR PLANE AREAS. 195 

whenever the given system has a pair of conjugate axes which 
can be located by inspection. 

214. Central Ellipse. — It is evident that an inertia-curve 
can be found with its center at any assumed point. That 
eUipse whose center is the centroid of the given system is called 
the central ellipse for the system. 

Since the central ellipse gives at once the radius of gyration 
for every axis through the centroid of the system, it enables us 
to determine readily the radius of gyration for any axis what- 
ever, by means of the known relation between radii of gyration 
for parallel axes. (Art. 196.) 

§ 3. Inertia-Curves for Plane Areas. 

215. General Principles. — The principles deduced in the 
treatment of inertia-curves for systems of forces are all true for 
the case of plane areas. But special difficulties arise in dealing 
with areas, because of the fact that the system of forces equiva- 
lent to any area consists of an infinite number of forces. The 
principles already developed are, however, sufficient to deal at 
least approximately with all areas, and accurately with many 
cases. 

216. Inertia-Curve an Ellipse. — Since the forces conceived 
to replace the elements of area (Art. 198) have all the same 
sign, the value of k'^ is always positive, and the inertia-curve is 
always an ellipse. (Arts. 209 and 211.) 

217. Cases Admitting Simple Treatment. — Whenever a pair 
of conjugate diameters can be located, and the radius of gyra- 
tion determined for each, the inertia-ellipse can be at once 
drawn as in Art. 213. This will be the case whenever it is 
possible to locate readily a pair of axes for which the product of 
inertia is zero. 

(i) If there is an axis of symmetry, this and any line perpen- 
dicular to it are a pair of conjugate axes (and in fact the principal 
axes) of the inertia-ellipse whose center is at their intersection. 
(Art. 200.) 



196 



GRAPHIC STATICS. 



(2) If two axes can be located in such a way that for every 
element of area whose distances from the axes are/, q, there is 
an equal element whose distances are /, —q, or — /, q, the 
product of inertia is zero for the two axes, and these are there- 
fore a pair of conjugate axes of the inertia-ellipse whose center 
is at their intersection. This of course includes the case when 
there is an axis of symmetry. (Art. 200.) 

When a pair of conjugate axes is known, the radius of gyration 
is to be found by one of the methods of Art. 202 or Art. 203 ; 
the ellipse can then be drawn exactly as explained in Art. 213. 

If a pair of conjugate axes cannot be located by inspection, 
the inertia-ellipse cannot be so readily constructed. Such cases 
will not be here treated. 

As examples of areas, in which the principal axes of the 
inertia-curve can be located by inspection, may be mentioned 
the cross-section of the I-beam, the deck-beam, the channel-bar, 
and other shapes of structural iron and steel. 

Many geometrical figures possess axes of symmetry. In 
others a pair of conjugate axes can be located by principle (2). 
Some of these will be discussed in the next article. 



Example. — Draw the central ellipse for the deck-beam sec- 
tion shown in Fig. 74. 



US 



IT 



"^2 



[Suggestions. — Since there is an axis of symmetry, this 
contains one of the principal axes of the eUipse. The other 
can be drawn as soon as the centroid of the section is 
known. Find the radius of gyration for each axis by 
Art. 202, and then construct the ellipse as explained in 
Art. 213.] 

2 1 8. Central Ellipses for Geometrical Fig- 
ures. — In many of the simple geometrical 
figures, not only can a pair of conjugate 
axes be located by inspection, but the radius 
of gyration for each of these axes can be found by a simple 
construction, so that the central ellipse can be readily drawn. 
Some of these cases will be here summarized. 




ITig. 74 



INERTIA-CURVES FOR PLANE AREAS. 



197 



(i) Parallelogram. — Let ABCD (Fig. 75) be the parallelo- 
gram ; then XX\ YV, drawn through the centroid parallel to 
the sides, are a pair of 
conjugate axes of the cen- A 

tral ellipse. Let AB = b, 
BC=d, and let h— the 
perpendicular distance be- 
tween AB and DC. The 
moment of inertia of the 
parallelogram with re- 
spect to the axis XX' is 
■equal to the moment of inertia of a rectangle of sides b and h. 
Hence k^, the square of the radius of gyration for this axis, is 

— • The length of k can be found by the construction used in 
12 

case of the rectangles in Fig. 70. The following modification 
of the method is, however, more convenient : 

Make EF=\BC, EG = \ BC, and draw a semicircle with FG 
as a diameter. From E draw a line perpendicular to BC, inter- 
secting the semicircle at /. Lay off EH=EI\ then a line 
through H parallel to XX^ is a tangent to the central ellipse. 
For by construction, 




EH=ET= 



BC 



12 



And since the projection of BC on a line perpendicular to XX is, 
equal to h, the projection of EH on the same line is equal to 



V 



12 



that is to k. 

The tangent parallel to the side BC may be found in a simi- 
lar way. It may, however, be located more simply as follows : 
It is evident that the distance between YY' and a tangent par- 
allel to it, measured along AB, bears the same ratio to AB that 
EH does to BC Hence, the parallelogram formed by the four 
tangents, two parallel to XX' and two parallel to YY', is simi- 
lar to the parallelogram ABCD. 




1^8 GRAPHIC STATICS. 

Fig. 75 shows this parallelogram and also the ellipse. 

(2) Triangle. — Let ABC (Fig. "jQ) be the triangle ; b = 
fy length of the base BC; d' = 

length of projection of BC on a 
line perpendicular to AD ; <^ = al- 
titude measured perpendicular to 
BC. AD and a line through the 
centroid parallel to BC are a pair 
of conjugate axes of the central 
ellipse (Art. 217). From Art. 
199, the radius of gyration for a 
central axis parallel to BC is 

\/ — = \/- • -. Let XX' be this axis, and H the point in which 
^18 ^3 6 ^ 

it intersects ^(S". TYiQXi HC=\AC. T^kQ HK=IAC=\HC, 
and make KC the. diameter of a semicircle. From Zf draw HI 
perpendicular to AC, intersecting the semicircle at /. Make 
HL=HI; then the line through L parallel to XX' is a tangent 
to the central ellipse. For the radius of gyration with respect 
to XX' is to HL as the altitude d is to AC. 

Again, for the axis AD, the radius of gyration is "V/ — ■ (Art. 

24 
199). Make DE=^ BC and DF=\ BC, and take EF as a 

diameter of a semicircle. From D draw a line perpendicular to 
BC, intersecting the semicircle in M; and make DG=DM; then 
a line from G parallel to AD is a tangent to the central ellipse. 
The figure shows the parallelogram formed by the two tan- 
gents parallel to XX' and the two parallel to AD, and also the 
central ellipse. 

(3) Ellipse. — From Art. 199, the radii of gyration of an 
ellipse with respect to the two principal diameters are ^a 
and |- b. Hence the central ellipse of inertia is similar to the 
given ellipse, its semi-axes being ^ a and ^b. A special case 
of this is a circle, for which the central curve is a circle whose 
radius is half that of the given circle. 



INERTIA-CURVES FOR PLANE AREAS. 



199 









Y 






x 


-^ 


C 


^ 


X' 


/ 


IV 


^ 


\ X 




1 




J 


\ 








1 




A E 


D 


B 








Y' 





(4) Semicircle. — Let ABC (Fig. 'j'j) be the semicircle, O 

being the centroid. (The point O may be located by the 

method described in Art. 184.) 

„ ^ .. . . 1 , Fig, rr 

Prom symmetry it is evident 

that the principal axes of the 
central ellipse are XX' and 
YY^, drawn through O, re- 
spectively parallel and per- 
pendicular to AB. 

With respect to the axis 
YY\ the radius of gyration 
is evidently |- r, the same as 

for the whole circle. Hence two lines parallel to YY^ and dis- 
tant \ r from it are tangents to the central ellipse. 

Again, the radius of gyration of the semicircle with respect 
to AB as an axis is also \ r, the same as for the whole circle.. 
To find it for the axis XX , with Z) as a center, and radius \ r,. 
draw an arc intersecting XX at F\ then OF = DF ~0D . 
But DF'is equal to the radius of gyration with respect to AB,. 
and ODis the distance between XX' and AB ; hence (Art. 196) 
OF is equal to the radius of gyration with respect to XX'. 
Hence if two lines are drawn parallel to XX' , each at a distance 
from it equal to OF, they will be tangents to the central ellipse. 
The ellipse can now be drawn in the usual manner. 



219. Summary of Results. — By the principles and methods 
developed in the present chapter, inertia-curves can be drawn 
for all the simpler cases that may arise ; namely, whenever a. 
pair of conjugate axes can be located by inspection. This will 
be the case whenever the product of inertia can be seen to be 
zero for any pair of axes ; and it includes every case of an area 
possessing an axis of symmetry. 

It is believed that this chapter contains as complete a discus- 
sion as is needed by the student of engineering. Those who- 
desire to pursue the subject further may consult other works. 





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Pig-. 50 



THE ELEMENTS OF PHYSICS. 



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Professor of Physics in Cornell University, 



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WITH NUMEROUS ILLUSTRATIONS. 

5 Vol. I. Mechanics and Heat. 
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III. Sound and Light. 

Price $1.50, net, per volume. 



It has been written with a view to providing a text-book which shall correspond with 
-the increasing strength of the mathematical teaching in our university classes. In most of 
the existing text-books it appears to have been assumed that the student possesses so 
scanty a mathematical knowledge that he cannot understand the natural language of 
physics, i.e., the language of the calculus. Some authors, on the other hand, have assumed 
a degree of mathematical training such that their work is unreadable for nearly all under- 
graduates. 

The present writers having had occasion to teach large classes;, the members of which 
were acquainted with the elementary principles of the calculus, have sorely felt the need of 
a text-book adapted to their students. The present work is an attempt on their part to 
supply this want. It is believed that in very many institutions a similar condition of affairs 
exists, and that there is a demand for a work of a grade intermediate between that of the 
existing elementary texts and the advanced manuals of physics. 

No attempt has been made in this work to produce a complete manual or compendium 
of experimental physics. The book is planned to be used in connection with illustrated 
lectures, in the course of which the phenomena are demonstrated and described. The 
authors have accordingly confined themselves to a statement of principles, leaving the 
lecturer to bring to notice the phenomena based upon them. In stating these principles, 
free use has been made of the calculus, but no demand has been made upon the student 
beyond that supplied by the ordinary elementary college courses on this subject. 

Certain parts of physics contain real and unavoidable difficulties. These have not been 
slurred over, nor have those portions of the subject which contain them been omitted. It 
has been thought more serviceable to the student and to the teacher who may have occa- 
sion to use the book to face such difficulties frankly, reducing the statements involving 
them to the simplest form which is compatible with accuracy. 

In a word, the Elements of Physics is a book which has been written for use in such 
institutions as give their undergraduates a reasonably good mathematical training. It is 
intended for teachers who desire to treat their subject as an exact science, and who are 
prepared to supplement the brief subject-matter of the text by demonstration, illustration, 
and discussion drawn from the fund of their own knowledge. 



THE MACMILLAN COMPANY. 

NEW YORK: CHICAGO: 

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A Laboratory flanual 

OF 

Physics and Applied Electricity. 

ARRANGED AND EDITED BY 

EDWARD L. NICHOLS, 

Professor of Physics in Cornell University. 
IN TWO VOLUMES. 

Vol L JUNIOR COURSE IN GENERAL PHYSICS. 

BY 

ERNEST MERRITT and FREDERICK J. ROGERS. 

Cloth. $3.00. 

VoL n. SENIOR COURSES AND OUTLINE OF 
ADVANCED WORK. 



GEORGE S. MOLER, FREDERICK BEDELL, HOMER J. HOTCHKISS, 
CHARLES P. MATTHEWS, and THE EDITOR. 

Cloth, pp. 444. $3.25. 



The first volume, intended for beginners, affords explicit directions adapted to a modern 
laboratory, together with demonstrations and elementary statements of principles. It is 
assumed that the student possesses some knowledge of analytical geometry and of the cal- 
culus. In the second volume more is left to the individual effort and to the maturer intel- 
ligence of the practicant. 

A large proportion of the students for whom primarily this Manual is intended, are pre- 
paring to become engineers, and especial attention has been devoted to the needs of that 
class of readers. In Vol. II., especially, a considerable amount of work in applied elec- 
tricity, in photometry, and in heat has been introduced. 

COMMENTS. 

" The work as a whole cannot be too highly commended. Its brief outlines of the 
various experiments are very satisfactory, its descriptions of apparatus are excellent ; its 
numerous suggestions are calculated to develop the thinking and reasoning powers of the 
student. The diagrams are carefully prepared, and its frequent citations of original 
sources of information are of the greatest value." — Street Railway Journal. 

" The work is clearly and concisely written, the fact that it is edited by Professor Nichols 
being a sufficient guarantee of merit." — Electrical Etigmeering. 

" It will be a great aid to students. The notes of experiments and problems reveal 
much original work, and the book will be sure to commend itself to instructors." 

^ San Francisco Chronicle. 



THE MACMILLAN COMPANY, 

NEW YORK: CHICAGO: 

66 FIFTH AVENUE. ROOM 23, AUDITORIUM. 



A LABORATORY MANUAL 



PHYSICS AND APPLIED ELECTRICITY. 

1 , ARRANGED AND EDITED BY 

EDWARD L. NICHOLS. 



COMMENTS. 



The work as a whole cannot be too highly commended. Its brief outlines of the 
various experiments are very satisfactory, its descriptions of apparatus are excellent ; 
its numerous suggestions are calculated to develop the thinking and reasoning powers 
of the student. The diagrams are carefully prepared, and its frequent citations of 
original sources of information are of the greatest value. — Street Railway Journal. 

The work is clearly and concisely written, the fact that it is edited by Professor 
Nichols being a sufficient guarantee of merit. — Electj'ical Efigineering. 

It will be a great aid to students. The notes of experiments and problems reveal 
much original work, and the book will be sure to commend itself to instructors. 

— S. F. Chronicle. 



Immediately upon its publication, Nichols' Laboratory Manual of 
Physics and Applied Electricity became the i-equired text-book in 
the following colleges, among others : Cornell University ; Princeton 
College ; University of Wisconsin ; University of Illinois ; Tulane 
University ; Union University, Schenectady, N.Y. ; Alabama Poly- 
technic Institute ; Pennsylvania State College ; Vanderbilt Uni- 
versity ; University of Nebraska ; Brooklyn Polytechnic Institute ; 
Maine State College ; Hamilton College, Clinton, N.Y. ; Wellesley 
College ; Mt. Holyoke College ; etc., etc. 

It is used as a reference manual in many other colleges where the 
arrangement of the courses in Physics does not permit its formall 
introduction. 



THE MACMILLAN COMPANY. 

NEW YORK: CHICAGO: 

66 FIFTH AVENUE. ROOM 23, AUDITORIUM. 



WORKS ON PHYSICS. 



A TEXT-BOOK OF THE PRINCIPLES OF PHYSICS. 

By ALFRED DANIELL, F.R.5.E., 

Late Lecturer on Physics in the School of Medicine, Edinburgh. 

Third Edition. Illustrated. 8vo. Cloth. Price, $4.cx). 

" I have carefully examined the book and am greatly pleased with it. It seemed to me a particu- 
larly valuable reference-book for teachers of elementary physics, as they would find in it many sugges- 
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principles of the science. I consider the book in its recent form a distinct advance in the text-books 
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reference-book." — Prof. Herbert T. Wade, Department of Physics, Columbia College. 

A LABORATORY MANUAL OF EXPERIMENTAL PHYSICS. 

By W. J. LOUDON AND J. C. HcLENNAN, 

Demonstrators in Physics, University of Toronto. 

Cloth. 8vo. pp. 302. Price, $1.90, jiet. 

The book contains a series of elementary experiments specially adapted for students who have had 
but little acquaintance with higher mathematical methods; these are arranged, as far as possible, in 
order of difficulty. There is also an advanced course of experimental work in Acoustics, Heat, and 
' Electricity and Magnetism, which is intended for those who have taken the elementary course. 

WORKS BY R. T. GLAZEBROOK, M.A. 

.MECHANICS AND HYDROSTATICS. Containing: I. Dynamics. II. Statics. III. Hydro- 
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•STATICS, pp.180. Price, 90 cents, «<?if. DYNAMICS, pp.256. Price, $1.25, ?2^/. 

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BY S. L. LONEY, 

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ELEMENTS OF STATICS AND DYNAMICS. 
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— Benjamin W. Snow, Professor of Physics, Indiana University. 

MECHANICS AND HYDROSTATICS FOR BEGINNERS. 

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A TREATISE ON ELEMENTARY DYNAMICS. 

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